If $$\displaystyle I=\int e^{-x}\log (e^{x}+1)dx$$, then I equals

$$\displaystyle x-(e^{-x}+1)\log (e^{x}+1)+C$$

$$\displaystyle x+(e^{-x}+1)\log (e^{x}+1)+C$$

$$\displaystyle x+(e^{x}+1)\log (e^{x}+1)+C$$

If $$I$$ $$=\displaystyle \int \sqrt {\frac {5-x}{2+x}}dx$$, then $$I$$ equals

$$\sqrt {x+2}\sqrt {5-x}+3 \sin^{-1}\sqrt {\dfrac {x+2}{3}}+C$$

$$\sqrt {x+2}\sqrt {5-x}+3 \sin^{-1}\sqrt {\dfrac {x+2}{7}}+C$$

$$\sqrt {x+2}\sqrt {5-x}+5 \sin^{-1}\sqrt {\dfrac {x+2}{5}}+C$$

Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number $$n$$. If $$\displaystyle \int \frac {dx}{x^n \sqrt {ax + b}} = -\frac {\sqrt {ax + b}}{(n - 1) bx^{n - 1}} - A \int \frac {dx}{x^{n - 1} \sqrt {ax + b}} + C$$, then $$A$$ is equal to

$$\displaystyle \frac {2n - 3}{2n - 2} \frac {a}{b}$$

$$\displaystyle \frac {2n - 3}{2n - 1}$$

$$\displaystyle \frac {n - 2}{n - 1} \frac {a}{b}$$

$$\displaystyle \frac {n - 3}{n - 2} \frac {a}{b}$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 12

$\displaystyle I=\int e^{-x}\log (e^{x}+1)dx$ , then I equals

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$\displaystyle x+(e^{-x}+1)\log (e^{x}+1)+C$
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$\displaystyle x+(e^{x}+1)\log (e^{x}+1)+C$
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$\displaystyle x-(e^{-x}+1)\log (e^{x}+1)+C$
0%
none of these

Explanation

$\displaystyle I=\int e^{-x}\log (e^{x}+1)dx$
Applying integration by parts

$\displaystyle I=-e^{-x}\log (e^{x}+1)+\int \frac{e^{-x}e^{x}}{e^{x}+1}dx$

$\displaystyle I=-e^{-x}\log (e^{x}+1)+\int \frac{1}{e^{x}+1}dx$

$=-\displaystyle e^{-x}\log (e^{x}+1)+\int \frac{e^{-x}}{e^{-x}+1}dx$

Put

$e^{-x}+1=t$ in the integral

$-e^{-x}dx=dt$

$\displaystyle I =-e^{ -x }\log (e^{ x }+1)-\int \frac { 1 }{ t } dt$

$=-\displaystyle e^{-x}\log (e^{x}+1)-\log (e^{x}+1)+C$

$=-\displaystyle e^{-x}\log (e^{x}+1)-\log (1+e^{x})+x+C$

$=-\displaystyle (e^{-x}+1)\log (e^{x}+1)+x+C$

$I$

$=\displaystyle \int \sqrt {\frac {5-x}{2+x}}dx$ , then

$I$ equals

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$\sqrt {x+2}\sqrt {5-x}+3 \sin^{-1}\sqrt {\dfrac {x+2}{3}}+C$
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$\sqrt {x+2}\sqrt {5-x}+3 \sin^{-1}\sqrt {\dfrac {x+2}{7}}+C$
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$\sqrt {x+2}\sqrt {5-x}+5 \sin^{-1}\sqrt {\dfrac {x+2}{5}}+C$
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None of these

$\int \cos(\ln x)dx=$

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$\dfrac{x}{2}(\cos \ln x+\sin \ln x)$
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$\dfrac{x}{2}[\cos \ln x-\sin \ln x]$
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$[x \cos \ln x + \sin \ln x]$
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None of these

Explanation

Let $I= \int \cos (\ln x) dx$

$\displaystyle I= x \cos (\ln x) + \int x \sin (\ln x) \frac{1}{x}dx$

$\Rightarrow I= x \cos (\ln x) + \int \sin (\ln x) dx$

$\displaystyle=x \cos (\ln x) + [x \sin (\ln x)-\int x \cos (\ln x).\frac{1}{x}dx]$

$\Rightarrow 2I=x(\cos (\ln x)+\sin (\ln x))$

$\therefore \displaystyle I=\frac{x}{2} (\cos (\ln x)+\sin (\ln x))+c$

The evaluation of

$\displaystyle \int \frac{pX^{p+2q-1}-qX^{q-1}}{X^{2p+2q}+2X^{p+q}+1}dx$ is

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$-\displaystyle \frac{X^p}{X^{p+q}+1}+C$
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$\displaystyle \frac{X^q}{X^{p+q}+1}+C$
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$-\displaystyle \frac{X^q}{X^{p+q}+1}+C$
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$\displaystyle \frac{X^p}{X^{p+q}+1}+C$

Explanation

Let us take

$\displaystyle \frac{x^{m}}{x^{(p+q)}+1}=t$ where m is either p or q as seen from the options,

$\displaystyle \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{(x^{(p+q)}+1)mx^{m-1}-x^{m}(p+q)x^{p+q-1}}{x^{2p+2q}+1+2x^{p+q}}$

$\displaystyle =\frac{(mx^{(p+q+m-1)}+mx^{m-1}-px^{p+q+m-1}-qx^{p+q+m-1})}{x^{2p+2q}+1+2x^{p+q}}$
we want powers of

$p+2q-1$ and

$q-1$ , so

$m=q,$

$\displaystyle =\frac{(qx^{(p+2q-1)}+qx^{q-1}-px^{p+2q-1}-qx^{p+2q-1})}{x^{2p+2q}+1+2x^{p+q}}$

$\displaystyle =\frac{qx^{q-1}-px^{p+2q-1}}{x^{2p+2q}+1+2x^{p+q}}$

$\displaystyle =\frac{\mathrm{d} t}{\mathrm{d} x}$
so the integral will be

$-\dfrac{x^{q}}{x^{(p+q)}+1}$

$\displaystyle \int x\sin x\sec ^{3}x dx$ is equal to

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$\displaystyle \frac{1}{2}\left [ \sec^{2}x-\tan x \right ]+c$
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$\displaystyle \frac{1}{2}\left [ x\sec^{2}x-\tan x \right ]+c$
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$\displaystyle \frac{1}{2}\left [ x\sec^{2}x+\tan x \right ]+c$
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$\displaystyle \frac{1}{2}\left [\sec^{2}x+\tan x \right ]+c$

Explanation

$\displaystyle \int x\sin x\sec ^{3}xdx$

$=\displaystyle \int x\sin x\frac{1}{\cos ^{3}x}dx$

$=\displaystyle \int x\tan x\sec ^{2}xdx$

$=\displaystyle x\int \sec x(\sec x\tan x)dx-\int \left [ \int\sec x(\sec x\tan x)dx \right ]dx+C$

Here , we will find the value of

$\int \sec x(\sec x\tan x)dx$
Put

$\sec x=t$

$\Rightarrow \sec x \tan x dx=dt$

$\displaystyle \int \sec x(\sec x\tan x)dx=\int t dt =\frac{t^{2}}{2}=\frac{\sec^{2}x}{2}$

$\therefore \displaystyle \int x\sin x\sec ^{3}xdx$

$=\displaystyle x\frac{\sec ^{2}x}{2}-\int \frac{\sec ^{2}x}{2}dx+C$

$=\displaystyle x\frac{\sec ^{2}x}{2}-\frac{\tan x}{2}+C$

$\displaystyle \int \frac{x^{2}\left ( x\sec^{2}x+\tan x \right )}{\left ( x\tan x+1 \right )^{2}}dx$

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$\displaystyle x^{2}\left [ -\frac{1}{x\tan x+1} \right ]+2\log \left( x\sin x+\cos x \right)+C$
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$\displaystyle x^{2}\left [ \frac{1}{x\tan x+1} \right ] +2\log \left( \sin x+x\cos x \right) +C$
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$\displaystyle x\left [ -\frac{1}{x\tan x+1} \right ]+2\log \left( \sin x+x\cos x \right)+C$
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$\displaystyle x\left [ \frac{1}{x\tan x+1} \right ] +2\log \left( x\sin x+\cos x \right) +C$

Explanation

$\displaystyle I = \int \frac{x^{2}\left ( x\sec^{2}x+\tan x \right )}{\left ( x\tan x+1 \right )^{2}}dx$
Note that d.c. of

$x \tan x+1$ is

$\displaystyle x\sec ^{2}x+\tan x.$
Hence integrating by parts, we get

$\displaystyle x^{2}\left [ -\frac{1}{x\tan x+1} \right ]+2\int \frac{x}{x\tan x+1}dx$
Now write

$\displaystyle I_1=2 \int \frac{x\cos x}{x\sin x+\cos x}$

$\displaystyle =2 \log \left ( x\sin x+\cos x \right )$

$\displaystyle \because$ d.c. of

$\displaystyle x\sin x+\cos x= x\cos x$
Thus

$\displaystyle I = x^{2}\left [ -\frac{1}{x\tan x+1} \right ]+2 \log \left ( x\sin x+\cos x \right )+C$

Integrate

$\displaystyle x\sqrt{{a^{2}-x^{2}} }$ with respect to

$x$ .

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$-(a^{2}-x^{2})+C$
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$\displaystyle\sqrt{a^2-x^2}+C$
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$-\displaystyle \dfrac{\left(a^2-x^2\right)^{\frac{3}{2}}}{3}+C$
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```
None of the above
```

$\displaystyle\int{\frac{e^{\displaystyle(x^2+4\ln{x})}-x^3e^{\displaystyle x^2}}{x-1}dx}$ is equal to

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$\displaystyle\left(\frac{e^{\displaystyle 2\ln{x}}-1}{2}\right)e^{\displaystyle x^2}+C$
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$\displaystyle\frac{(x-1)xe^{\displaystyle x^2}}{2}+C$
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$\displaystyle\frac{(x^2-1)}{2x}-e^{\displaystyle x^2}+C$
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none of these

Explanation

$\displaystyle \int \frac{e^{(x^{2}+4\ln x)}-x^{3}e^{x^{2}}}{x-1}$

$\displaystyle =\int \frac{e^{x^{2}}x^{4}-x^{3}e^{x^{2}}}{x-1}$

Let,

$x^{2}=t$

$2xdx=dt$

So,

$\displaystyle =\int \frac{e^{t}.x^{4}-x^{3}e^{t}}{x-1} dx$

$\displaystyle =\int \frac{e^{t}(x^{4}-x^{3})}{x-1} dx$

$\displaystyle =\int e^{t}x^{3} dx$

$=\int e^{t}x^{2}.xdx$

$=\int \dfrac{e^{t}}{2}.t dt$

$= \dfrac{1}{2}(t.e^{t}-\int e^{t}+c)$

$=\dfrac{1}{2} (e^{t}(t-1))+c$

$=\dfrac{1}{2}e^{x^{2}}(x^{2}-1)+c$

$\displaystyle \int \sin 2x.\log \cos x\: dx$ is equal to

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$\displaystyle \cos ^{2}x\left ( \frac{1}{2}+\log \cos x \right )+k$
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$\displaystyle \cos ^{2}x.\log \cos \: x+k$
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$\displaystyle \cos ^{2}x\left ( \frac{1}{2}-\log \cos x \right )+k$
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none of these

Explanation

$\displaystyle \int \sin 2x.\log \cos x\: dx$

$\displaystyle =\log \cos x\: (-\frac { \cos { 2x } }{ 2 } )-\int { \frac { -\sin { x } }{ \cos { x } } } (-\frac { \cos { 2x } }{ 2 } )dx$

$\displaystyle =-\frac { 1 }{ 2 } (2\cos ^{ 2 }{ x } -1)\log \cos x\: -\frac { 1 }{ 2 } \int { \frac { \sin { x } }{ \cos { x } } } (2\cos ^{ 2 }{ x } -1)dx$

$\displaystyle =-\cos ^{ 2 }{ x } \log \cos x\: +\frac { 1 }{ 2 } \log \cos x-\int { \sin { x\cos { x } } dx } +\frac { 1 }{ 2 } \int { \tan { x } } dx$

Put

$\cos x=t$

$\Rightarrow -\sin xdx=dt$

$\displaystyle =-\cos ^{ 2 }{ x } \log \cos x\: +\frac { 1 }{ 2 } \log \cos x+\int { tdt } -\frac { 1 }{ 2 } \log \cos x$

$\displaystyle =-\cos ^{ 2 }{ x } \log \cos x\: +\frac { { t }^{ 2 } }{ 2 } +C$

$\displaystyle =\cos ^{2}x\left ( \frac{1}{2}-\log \cos x \right )+C$

$\displaystyle \int { f(x) } dx=F(x)$ , then

$\displaystyle \int { { x }^{ 3 } } f\left( { x }^{ 2 } \right) dx$ equals to

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$\displaystyle \frac {1}{2} \left[x^2(F(x))^2 - \int (F(x))^2 dx \right]$
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$\displaystyle \frac {1}{2} \left[x^2F(x^2) - \int F(x^2) d(x^2) \right]$
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$\displaystyle \frac {1}{2} \left[x^2F(x) - \frac{1}{2} \int (F(x))^2 dx \right]$
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$\displaystyle \frac {1}{2} \left[x^2F(x^2) + \int F(x^2) d(x^2) \right]$

Explanation

Given

$\displaystyle \int { f(x) } dx=F(x)$

Let

$I=\displaystyle \int { { x }^{ 3 } } f\left( { x }^{ 2 } \right) dx$

Put

$x^{ 2 }=t$

$\Rightarrow \displaystyle xdx=\frac { dt }{ 2 }$

$\Rightarrow \displaystyle I=\frac { 1 }{ 2 } \int tf(t)dt$

$=\displaystyle \frac { 1 }{ 2 } \left[t\int f(t)dt-\int \left(\int { f(t)dt } \right)dt \right]$

$\left[ \text{Integration by parts}\displaystyle \int {(u\cdot v)} dx= u\int v dx-\int{\left(\dfrac{du}{dx}\int v dx \right)}dx\right]$

$\Rightarrow \displaystyle I=\frac { 1 }{ 2 } [tF(t)-\int F(t)dt]$

$\Rightarrow \displaystyle I=\frac { 1 }{ 2 } [x^{ 2 }F(x^{ 2 })-\int F(x^{ 2 })d(x^{ 2 })]$

$\displaystyle \int x\log \left ( 1+x^{2} \right )dx=\phi \left ( x \right ).\log \left ( 1+x^{2} \right )+\Psi \left ( x \right )+c$ then

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$\displaystyle \phi \left ( x \right )=\frac{1+x^{2}}{2}$
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$\displaystyle \Psi \left ( x \right )=\frac{1+x^{2}}{2}$
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$\displaystyle \Psi \left ( x \right )=-\frac{1+x^{2}}{2}$
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$\displaystyle \phi \left ( x \right )=-\frac{1+x^{2}}{2}$

Explanation

Given,

$\int x\log \left( 1+x^{ 2 } \right) dx=\phi \left( x \right) .\log \left( 1+x^{ 2 } \right) +\Psi \left( x \right) +c$

Consider,

$\int x\log \left( 1+x^{ 2 } \right) dx$

$\displaystyle =\log \left( 1+x^{ 2 } \right) \frac { x^{ 2 } }{ 2 } -\int { (\frac { 2x }{ 1+x^{ 2 } } } )\frac { x^{ 2 } }{ 2 } dx$

$\displaystyle =\log \left( 1+x^{ 2 } \right) \frac { x^{ 2 } }{ 2 } -\int { \frac { x^{ 3 } }{ 1+x^{ 2 } } } dx$

$\displaystyle =\frac { x^{ 2 } }{ 2 } \log \left( 1+x^{ 2 } \right) +\int { (-x+\frac { x }{ x^{ 2 }+1 } ) } dx$

$\displaystyle =\frac { x^{ 2 } }{ 2 } \log \left( 1+x^{ 2 } \right) -\frac { x^{ 2 } }{ 2 } +\int { \frac { x }{ x^{ 2 }+1 } } dx$

Substitute

$x^{ 2 }+1=t$

$\Rightarrow xdx=\dfrac { dt }{ 2 }$

$\displaystyle=\frac { x^{ 2 } }{ 2 } \log \left( 1+x^{ 2 } \right) -\frac { x^{ 2 } }{ 2 } +\frac { 1 }{ 2 } \int { \frac { 1 }{ t } } dt$

$\displaystyle =\frac { x^{ 2 } }{ 2 } \log \left( 1+x^{ 2 } \right) -\frac { x^{ 2 } }{ 2 } +\frac { 1 }{ 2 } \log \left( 1+x^{ 2 } \right) +C$

$\displaystyle =\frac { x^{ 2 }+1 }{ 2 } \log \left( 1+x^{ 2 } \right) -\frac { x^{ 2 }+1 }{ 2 } +C'$
On comparing with (1), we get

$\displaystyle \phi (x)=\frac { x^{ 2 }+1 }{ 2 } ,\psi (x)=-\frac { x^{ 2 }+1 }{ 2 }$

$\displaystyle \int { \left( \sqrt { { x }^{ 2 }+1 } \right) \left[ \frac { \ln { \left( { x }^{ 2 }+1 \right) } -2\ln { x } }{ { x }^{ 4 } } \right] dx } ;g\left( x \right)=\left( 1+\frac { 1 }{ { x }^{ 2 } } \right) ;$ equals

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$\displaystyle \frac { 1 }{ 3 } { \left( g\left( x \right) \right) }^{ 1/2 }\ln { \left( g\left( x \right) \right) } +\frac { 2 }{ 9 } \left( g\left( x \right) \right) +C$
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$\displaystyle -\frac { 1 }{ 3 } { \left( g\left( x \right) \right) }^{ 3/2 }\ln { \left( g\left( x \right) \right) } +\frac { 2 }{ 9 } { \left( g\left( x \right) \right) }^{ 3/2 }+C$
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$\displaystyle \frac { 1 }{ 3 } { \left( g\left( x \right) \right) }^{ 3/2 }\ln { \left( g\left( x \right) \right) } -\frac { 2 }{ 9 } \left( g\left( x \right) \right) +C$
0%
None of these

Explanation

Let

$\displaystyle I=\int { \left( \sqrt { { x }^{ 2 }+1 } \right) \left[ \frac { \ln { \left( { x }^{ 2 }+1 \right) } -2\ln { x } }{ { x }^{ 4 } } \right] dx }$

$\displaystyle =\int { x\left( \sqrt { 1+\frac { 1 }{ { x }^{ 2 } } } \right) \frac { { \left[ \ln { { x }^{ 2 } } +\ln { \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) } -\ln { { x }^{ 2 } } \right] }^{ 2 } }{ { x }^{ 4 } } dx }$

$\displaystyle =\int { x\left( \sqrt { 1+\frac { 1 }{ { x }^{ 2 } } } \right) \frac { { \left[ \ln { { x }^{ 2 } } +\ln { \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) } -2\ln { { x } } \right] }^{ 2 } }{ { x }^{ 4 } } dx }$

$\displaystyle =\int { \left( \frac { 1 }{ { x }^{ 3 } } \sqrt { 1+\frac { 1 }{ { x }^{ 2 } } } \left[ \ln { \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) } \right] \right) dx }$

Putting

$\displaystyle 1+\frac { 1 }{ { x }^{ 2 } } =t\Rightarrow \frac { -2 }{ { x }^{ 3 } } dx=dt,$ we get

$\displaystyle \Rightarrow I=\left( \frac { -1 }{ 2 } \right) \int { \sqrt { t } \ln { t } dt } =\left( \frac { -1 }{ 2 } \right) \left[ \int { { t }^{ 1/2 }.\ln { t } dt } \right]$

$\displaystyle =-\frac { 1 }{ 2 } \int { { t }^{ 1/2 }.\ln { t } dt } =-\frac { 1 }{ 2 } \left[ \ln { t } \int { { t }^{ 1/2 }dt } -\int { \frac { 1 }{ t } \left( { t }^{ 1/2 }dt \right) dt } \right]$

$\displaystyle =-\frac { 1 }{ 2 } \left[ \left( \ln { t } \right) \frac { { t }^{ 3/2 } }{ 3/2 } -\int { \frac { { t }^{ 1/2 } }{ 3/2 } dt } \right] =-\frac { 1 }{ 3 } { t }^{ 3/2 }\ln { t } +\frac { 1 }{ 3 } .\frac { { t }^{ 3/2 } }{ 3/2 } +c$

$\displaystyle \therefore I=-\frac { 1 }{ 3 } { \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) }^{ 3/2 }\ln { \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) } +\frac { 2 }{ 9 } { \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) }^{ 3/2 }+C$

$\displaystyle =-\frac { 1 }{ 3 } { \left( g\left( x \right) \right) }^{ 3/2 }\ln { \left( g\left( x \right) \right) } +\frac { 2 }{ 9 } { \left( g\left( x \right) \right) }^{ 3/2 }+C$

$\displaystyle \int \frac{e^{\tan^{-1} x}}{(1+x^2)}\left [\left ( \sec^{-1} \sqrt{1+x ^2} \right )^2+ \cos^{-1} \left ( \frac{1-x^2}{1+x^2} \right )dx \right ]$ at

$(x>0)$

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$e^{\tan ^{-1}x}.\tan^{-1} x+C$
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$\displaystyle \frac{e^{\tan ^{-1}x}.\left ( \tan^{-1} x \right )^2}{2}+C$
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$e^{\tan ^{-1}x}.\left ( \sec^{-1} \left ( \sqrt{1+x^2} \right )\right )^2+C$
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$e^{\tan ^{-1}x}.\left ( \mathrm{cosec} ^{-1} \left ( \sqrt{1+x^2} \right )\right )^2+C$

Explanation

Let

$\int { \cfrac { { e }^{ { \tan }^{ -1 }(x) } }{ 1+{ x }^{ 2 } } } \left[ { \left( { \sec }^{ -1 }\sqrt { 1+{ x }^{ 2 } } \right) }^{ 2 }+{ \cos }^{ -1 }\left( \cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) \right] dx=I$

Substitute

${ \tan }^{ -1 }(x)=t$ and

$\cfrac { dx }{ 1+{ x }^{ 2 } } =dt$

We get,

$I=\int { { e }^{ t } } \left[ { \left( t \right) }^{ 2 }+2t \right] dt\\ =\int { { t }^{ 2 } } { e }^{ t }dt+\int { 2t{ e }^{ t }dt }$

By integration by parts we get

$I={ t }^{ 2 }{ e }^{ t }-\int { 2t{ e }^{ t }dt } +\int { 2t{ e }^{ t }dt } +C\\ { =t }^{ 2 }{ e }^{ t }+C\\ ={ e }^{ { \tan }^{ -1 }(x) }{ \left( { \sec }^{ -1 }\sqrt { 1+{ x }^{ 2 } } \right) }^{ 2 }+C$

Hence, option 'C' is correct.

$\displaystyle \int \left [{ \log \left( 1+\cos x \right) -x\tan \left( \frac{x}{2} \right) }\right ] dx$

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$\displaystyle x\log\left ( 1+\tan x \right )$
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$\displaystyle x\log\left ( 1+\sin x \right )$
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$\displaystyle \log\left ( 1+\sin x \right )$
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$\displaystyle x\log\left ( 1+\cos x \right )$

Explanation

$I = \displaystyle\int \log\left ( 1+\cos x \right )dx-\int x\tan\left ( x/2 \right )dx$

$\displaystyle I=x\log\left ( 1+\cos x \right )+\int x\frac{\sin x}{1+\cos x}dx-\int x\tan \frac{x}{2}$ .

$\displaystyle =x\log\left ( 1+\cos x \right )$ second term gets cancel

$\displaystyle \left[ \because \frac{\sin x}{1+\cos x}=\tan \frac{x}{2}\right ]$

$\displaystyle =x\log\left ( 1+\cos x \right )$

$\displaystyle\int \frac{2x+\sin 2x}{1+\cos 2x}dx$

0%
$x\cot x.$
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$x\tan x.$
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$x^{2}\tan x.$
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$x\sec x.$

Explanation

Let

$\displaystyle I=\int { \frac { 2x+\sin { 2x } }{ \cos { 2x } +1 } dx } =\int { \left( \frac { 2x }{ \cos { 2x } +1 } +\frac { \sin { 2x } }{ \cos { 2x } +1 } \right) dx }$

$\Rightarrow I={ I }_{ 1 }+{ I }_{ 2 }$
Where

$\displaystyle { I }_{ 1 }=\int { \frac { \sin { 2x } }{ \cos { 2x } +1 } dx }$

Put

$t=2x\Rightarrow dt=2dx$

$\displaystyle { I }_{ 1 }=\frac { 1 }{ 2 } \int { \frac { \sin { t } }{ \cos { t } +1 } dt }$

Now put

$u=\cos { t } +1\Rightarrow du=-\sin { t } dt$

$\displaystyle { I }_{ 1 }=-\frac { 1 }{ 2 } \int { \frac { du }{ u } } =-\frac { \log { u } }{ 2 }$

$\displaystyle =-\frac { \log { \left( \cos { t } +1 \right) } }{ 2 } -\frac { \log { \left( \cos { 2x } +1 \right) } }{ 2 }$

And

$\displaystyle { I }_{ 2 }=\int { \frac { 2x }{ \cos { 2x } +1 } dx } =\int { x\sec ^{ 2 }{ x } dx }$

$\displaystyle =x\tan { x } -\int { \tan { x } dx } =x\tan { x } +\log { \left( \cos { x } \right) }$

Hence

$\displaystyle I=-\frac { \log { \left( \cos { 2x } +1 \right) } }{ 2 } +x\tan { x } +\log { \left( \cos { x } \right) }$

$\displaystyle =x\tan { x } -\frac { 1 }{ 2 } \log { \left( 2\cos ^{ 2 }{ x } \right) } +\log { \left( \cos { x } \right) }$

$=x\tan { x } +c$
Hence, option 'B' is correct.

$\displaystyle I=\int_{0}^{1}\frac{e^{t}}{1+t}$ dt, then

$\displaystyle p=\int_{0}^{1}e^{t}\log\left ( 1+t \right )dt=$

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$I$
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$2I$
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$\displaystyle e\log2-I$
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none

Explanation

Given,

$\displaystyle p=\int_{0}^{1}e^{t}\log\left ( 1+t \right )dt$

Also,given,

$\displaystyle I=\int_{0}^{1}\frac{e^{t}}{1+t} dt$

$I={ [e^{ t }\log \left( 1+t \right) ] }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 } e^{ t }\log \left( 1+t \right) dt$

$\Rightarrow I=e\log 2-p$

$\therefore p=e \log2 -I$

Eavaluate :

$\displaystyle \int \displaystyle x\tan^{-1}x$

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$\displaystyle \frac{1}{2}(x^{2}+1)\tan ^{-1}x-\frac{1}{2}x+C$
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$\displaystyle \frac{1}{2}x^{2}\tan ^{-1}x-\frac{1}{2}x+ C$
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$\displaystyle \frac{1}{2}x^{2}\tan ^{-1}x+\frac{1}{2}x+C$
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none of these

Explanation

$\displaystyle I=\int x\tan ^{-1}x\:dx$

We know that,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$I\displaystyle =\left ( \tan ^{-1}x \right )\left ( \frac{x^{2}}{2} \right )-\int \frac{1}{1+x^{2}}\frac{x^{2}}{2}dx.$

$\displaystyle =\frac{1}{2}x^{2}\tan ^{-1}x-\frac{1}{2}\int \frac{x^{2}+1-1}{x^{2}+1}dx$

$\displaystyle =\frac{1}{2}x^{2}\tan ^{-1}x-\frac{1}{2}\int \left ( 1-\frac{1}{x^{2}+1} \right )dx$

$\displaystyle =\frac{1}{2}x^{2}\tan ^{-1}x-\frac{1}{2}\left [ x-\tan ^{-1}x \right ]$

$I\displaystyle =\frac{1}{2}(x^{2}+1)\tan ^{-1}x-\frac{1}{2}x+C$

$\displaystyle \int \left ( x^{3}-2x^{2}+3x-1 \right )\cos 2x\:dx$

$\displaystyle =\frac{\sin 2x}{4}u\left ( x \right )+\frac{\cos 2x}{8}v\left ( x \right )+C$ then

0%
$\displaystyle u\left ( x \right )=x^{3}-4x^{2}+3x$
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$\displaystyle u\left ( x \right )=2x^{3}-4x^{2}+3x$
0%
$\displaystyle u\left ( x \right )=3x^{3}-4x+3$
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$\displaystyle v\left ( x \right )=6x^{2}-8x$

Explanation

$\displaystyle \int \left ( x^{3}-2x^{2}+3x-1 \right )\cos 2x\:dx$

$\displaystyle =\frac{\sin 2x}{4}u\left ( x \right )+\frac{\cos 2x}{8}v\left ( x \right )+C$ ....

$(i)$

Applying the formula given in the comprehension, required integral becomes

$\displaystyle \int \left ( x^{3}-2x^{2}+3x-1 \right )\cos 2x\:dx$

$=\displaystyle \left (x^{3}-2x^{2}+3x-1 \right )\frac{\sin 2x}{2}-\left ( 3x^{2}-4x+3 \right )\left ( -\frac{\cos 2x}{4} \right )+\left ( 6x-4 \right )\left (-\dfrac{\sin 2x}{8} \right ) - 6\dfrac{\cos 2x}{16}+C$

$=\displaystyle \dfrac{\sin 2x}{4}(2x^{3}-4x^{2}+6x-2)-(3x-2)\dfrac{\sin 2x}{4}+\dfrac{\cos 2x}{8}(6x^{2}-8x+6)-3\dfrac{\cos 2x}{8}+C$

$\displaystyle =\frac{\sin 2x}{4}\left [ 2x^{3}-4x^{2}+3x \right ]+\frac{\cos 2x}{8}\left [ 6x^{2}-8x+3 \right ]+C$

Comparing with the given equation

$(i)$ , we get

$\displaystyle u\left ( x \right )=2x^{3}-4x^{2}+3x$

Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number

$n$ .

$\displaystyle \int \frac {dx}{x^n \sqrt {ax + b}} = -\frac {\sqrt {ax + b}}{(n - 1) bx^{n - 1}} - A \int \frac {dx}{x^{n - 1} \sqrt {ax + b}} + C$ , then

$A$ is equal to

0%
$\displaystyle \frac {2n - 3}{2n - 1}$
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$\displaystyle \frac {2n - 3}{2n - 2} \frac {a}{b}$
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$\displaystyle \frac {n - 2}{n - 1} \frac {a}{b}$
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$\displaystyle \frac {n - 3}{n - 2} \frac {a}{b}$

Explanation

$\displaystyle\int {\frac{dx}{ { x }^{ n }\left( \sqrt { ax+b } \right) } } = -\frac { \sqrt { ax+b } }{ \left( n-1 \right) b{ x }^{ n-1 } } -A\int { \frac { dx }{ { x }^{ n-1 }\left( \sqrt { ax+b } \right) } } + C$

$\Rightarrow\displaystyle \int { \frac { 1+Ax }{ { x }^{ n }\left( \sqrt { ax+b } \right) } } dx = -\frac { \sqrt { ax+b } }{ \left( n-1 \right) b{ x }^{ n-1 } } + C$

Differentiating on both sides,

$\dfrac{1+Ax}{x^n(\sqrt{ax+b})} =- \dfrac{x^{n-1}\dfrac{a}{2}(ax+b)^{-\frac12}-(ax+b)^{\frac12}(n-1)x^{n-2}}{b(n-1)x^{2n-2}}$

$\Rightarrow\dfrac{1+Ax}{x^n\sqrt{ax+b}}=-\dfrac{ax-2(n-1)(ax+b)}{b(n-1)2x^n\sqrt{ax+b}}$

Simplifying further, it gives

$A = \dfrac { 2n-3 }{ 2n-2 } \dfrac { a }{ b }$

$\displaystyle I=\int \sec^{2}x\:cosec ^{4}\:x\:dx= K\cot^{3}x+L\tan x+M\cot x+C$ then

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K $=-1/3$
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L $=2$
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M $=-2$
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none of these

Explanation

Let ,

$I=\int { \sec ^{ 2 }{ x\csc ^{ 4 }{ x } dx } }$

Using integration by parts ,

$I=\csc ^{ 4 }{ x\int { \sec ^{ 2 }{ x } dx } } -\int { \frac { d }{ dx } } \left( \csc ^{ 4 }{ x } \right) \int { \sec ^{ 2 }{ x } dx }$

$=\csc ^{ 4 }{ x\tan { x } } -\int { \left( -4\csc ^{ 4 }{ x\cot { x } } \right) } \tan { x } dx$

$=\csc ^{ 4 }{ x\tan { x } } +4\int { \csc ^{ 4 }{ x } } dx$

$=\csc ^{ 4 }{ x\tan { x } } +4\int { \csc ^{ 2 }{ x\csc ^{ 2 }{ x } } } dx$

applying integration by parts again.

$=\csc ^{ 4 }{ x\tan { x } } +4\left[ \csc ^{ 2 }{ x } \int { \csc ^{ 2 }{ x } dx-\int { \left( -2\csc ^{ 2 }{ x } \cot { x } \right) (-\cot { x } )dx } } \right]$

$=\csc ^{ 4 }{ x\tan { x } } +4\left[ \csc ^{ 2 }{ x(-\cot { x } )-2\int { \csc ^{ 2 }{ x } \cot ^{ 2 }{ x } dx } } \right]$

$=\csc ^{ 4 }{ x\tan { x } } -4\csc ^{ 2 }{ x\cot { x } +\frac { 8 }{ 3 } \cot ^{ 3 }{ x } }$

Substitute ,

$\csc ^{ 2 }{ x } =1+\cot ^{ 2 }{ x }$ and simplify

Then we get ,

$I=-\frac { 1 }{ 3 } \cot ^{ 3 }{ x } +\tan { x } -2\cot { x } +C$

Therefore ,

$K=-\frac { 1 }{ 3 }$

$L=1$

$M=-2$

$\displaystyle \int \frac {xe^x}{\sqrt {1 + e^x}} dx = f(x) \sqrt {1 + e^x} -2 \log \: g(x) + C$ , then

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$\displaystyle f(x) = x - 1$
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$\displaystyle g(x) = \frac {\sqrt {1 + e^x} - 1}{\sqrt {1 + e^x} + 1}$
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$\displaystyle g(x) = \frac {\sqrt {1 + e^x} + 1}{\sqrt {1 + e^x} - 1}$
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$\displaystyle f(x) = 2(x - 2)$

Explanation

Substituting

$1+{ e }^{ x }={ t }^{ 2 }\Rightarrow { e }^{ x }dx=2tdt,x=\log { \left( { t }^{ 2 }-1 \right) }$

$\int { \cfrac { x{ e }^{ x } }{ \sqrt { 1+{ e }^{ x } } } dx } \\ =\int { \left( \cfrac { \log { \left( { t }^{ 2 }-1 \right) } }{ t } \right) 2tdt } \\ =2\int { \log { \left( { t }^{ 2 }-1 \right) } dt } \\ =2\left[ t\log { \left( { t }^{ 2 }-1 \right) } -2\int { \cfrac { { t }^{ 2 } }{ { t }^{ 2 }-1 } dt } \right] +c\\ =2\left[ t\log { \left( { t }^{ 2 }-1 \right) } -2\left( 1+\cfrac { 1 }{ 2 } \log { \cfrac { t-1 }{ t+1 } } \right) \right] +c\\ =2x\sqrt { 1+{ e }^{ x } } -4\sqrt { 1+{ e }^{ x } } -2\log { \cfrac { \sqrt { 1+{ e }^{ x } } -1 }{ \sqrt { 1+{ e }^{ x } } +1 } } +c$

$\displaystyle \int \log (\sqrt {1 - x} + \sqrt {1 + x}) dx = xf(x) + Lx + M \: sin^{-1} x + C$ , then

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$\displaystyle f(x) = \log (\sqrt {1 - x} + \sqrt {1 + x})$
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$\displaystyle L = - \frac {1}{2}$
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$\displaystyle M = - \frac {2}{3}$
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$\displaystyle M = - \frac {1}{2}$

Explanation

Let

$\displaystyle I=\int { \log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } } dx$

$\displaystyle =x\log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } -\int { \frac { x }{ \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } } \left( \frac { -1 }{ 2\sqrt { 1-x } } +\frac { 1 }{ 2\sqrt { 1+x } } \right) dx$

$\displaystyle =x\log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } +\frac { 1 }{ 2 } { I }_{ 1 }$

Where

$\displaystyle { I }_{ 1 }=\int { \frac { \sqrt { 1+x } -\sqrt { 1-x } }{ \sqrt { 1+x } +\sqrt { 1-x } } } \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } dx$

$\displaystyle =\int { \frac { { \left( \sqrt { 1+x } -\sqrt { 1-x } \right) }^{ 2 } }{ 1+x-1+x } } \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } dx$

$\displaystyle =\frac { 1 }{ 2 } \int { \frac { \\ 1+x+1-x-2\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1-{ x }^{ 2 } } } } dx=\int { \left( \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } -1 \right) } dx=\sin ^{ -1 }{ x } -x$

Therefore

$\displaystyle I=x\log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } +\frac { 1 }{ 2 } \left( \sin ^{ -1 }{ x } -x \right) +c$

$\int (x^{3}\, -\, 2x^{2}\, +\, 3x\, -\, 1)\, cos2x\, dx\, =\, \displaystyle \frac{sin 2x}{4}u(x)\, +\, \frac{cos 2x}{8}v(x)\, +\, c$ , then

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$u(x)\, =\, x^{3}\, -\, 4x^{2}\, +\, 3x$
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$u(x)\, =\, 2x^{3}\, -\, 4x^{2}\, +\, 3x$
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$v(x)\, =\, 3x^{2}\, -\, 4x\, +\, 3$
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$v(x)\, =\,k 6x^{2}\, -\, 8x$

Explanation

$\int { \left( { x }^{ 2 }-2{ x }^{ 2 }+3x-1 \right) } \cos { 2x } dx\\ =\int { { x }^{ 2 } } \cos { 2x } dx-2\int { { x }^{ 2 }\cos { 2x } } dx+\int { 3x\cos { 2x } } dx-\int { \cos { 2x } } dx\\ =\frac { 1 }{ 2 } { x }^{ 3 }\sin { 2x } -\frac { 3 }{ 2 } \int { { x }^{ 2 } } \sin { 2x } dx-2\int { { x }^{ 2 }\cos { 2x } } dx+3\int { x } \cos { 2x } d-\int { \cos { 2x } dx }$

$\displaystyle =\frac { 3 }{ 4 } { x }^{ 2 }\cos { 2x } +\frac { 1 }{ 2 } { x }^{ 3 }\sin { 2x } +\frac { 3 }{ 2 } \int { x } \cos { 2x } dx-2\int { { x }^{ 2 }\cos { 2x } dx } -\int { \cos { 2x } } dx$

$\displaystyle ={ x }^{ 3 }\sin { x } \cos { x- } { x }^{ 2 }\sin { 2x } +\frac { 3 }{ 4 } { x }^{ 2 }\cos { 2x } \frac { 3 }{ 4 } x\sin { 2x } -x\cos { 2x } +\frac { 3 }{ 8 } \cos { 2x }$

$\displaystyle =\frac { 1 }{ 4 } \left( { 2x }^{ 3 }-{ 4x }^{ 2 }+3x \right) \sin { 2x } +\frac { 1 }{ 8 } \left( { 12x }^{ 3 }-{ 16x }^{ 2 }+6x \right) \cos { 2x }$

$\int \left [\displaystyle \frac {\sqrt{x^2\, +\, 1}\, [ln(x^2\, +\, 1)\, -\, 2\, ln\, x]}{x^4} \right ]$

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$\displaystyle \frac {(x^2\, -\, 1)\, \sqrt{x^2\, -\, 1}}{9x^3}\, \left [2\, -\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$
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$\displaystyle \frac {(x^2\, -\, 1)\, \sqrt{x^2\, -\, 1}}{9x^3}\, \left [2\, +\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$
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$\displaystyle \frac {(x^2\, +\, 1)\, \sqrt{x^2\, +\, 1}}{9x^3}\, \left [2\, -\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$
0%
$\displaystyle \frac {(x^2\, -\, 1)\, \sqrt{x^2\, +\, 1}}{9x^3}\, \left [2\, +\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$

Explanation

Let

$\displaystyle I=\int \frac { \sqrt { 1+x^{ 2 } } \left\{ \log \left( 1+x^{ 2 } \right) -2\log x \right\} }{ x^{ 4 } } dx$

$\displaystyle =\int \log \left\{ \frac { 1+x^{ 2 } }{ x^{ 2 } } \right\} \cdot \frac { \sqrt { 1+x^{ 2 } } }{ x^{ 1 }.x^{ 3 } } dx$

Substitute

$\displaystyle \frac { 1+x^{ 2 } }{ x^{ 2 } } =1+\frac { 1 }{ x^{ 2 } } =t\Rightarrow -\frac { 2 }{ x^{ 3 } } dx=dt$

$\displaystyle I=-\frac { 1 }{ 2 } \int \sqrt { t } .\log tdt=-\frac { 1 }{ 2 } \left[ \frac { 2 }{ 3 } t\sqrt { t } \log t-\frac { 4 }{ 9 } t\sqrt { t } \right]$

$\displaystyle =\frac { (x^{ 2 }\, +\, 1)\, \sqrt { x^{ 2 }\, +\, 1 } }{ 9x^{ 3 } } \, \left[ 2\, -\, 3\, ln\, \left( 1\, +\, \frac { 1 }{ x^{ 2 } } \right) \right]$

$\displaystyle I = \int \frac {\sec^{-1} \sqrt x - cosec^{-1} \sqrt x}{\sec^{-1} \sqrt x + cosec^{-1} \sqrt x} dx$ , then I equals

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$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) + x + C$
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$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x + \sqrt {x - 1}) - x + C$
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$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) - x + C$
0%
none of these

Explanation

Using

$\sec ^{ -1 } \sqrt { x } +cosec^{ -1 }\sqrt { x } =\cfrac { \pi }{ 2 }$ , we get

$I=\int { \cfrac { \sec ^{ -1 } \sqrt { x } -\cfrac { \pi }{ 2 } }{ \cfrac { \pi }{ 2 } } } dx=\cfrac { 4 }{ \pi } { I }_{ 1 }-x$
Where

${ I }_{ 1 }=\int { \sec ^{ -1 } \sqrt { x } dx }$
Putting

$\sqrt { x } =\sec { u }$ , we get

${ I }_{ 1 }=\int { u\left( 2\sec ^{ 2 }{ u } \tan { u } \right) du } \\ =u\sec ^{ 2 }{ u } -\int { \sec ^{ 2 }{ u } du } \\ =u\sec ^{ 2 }{ u } -\tan { u } \\ =x\sec ^{ -1 }{ \sqrt { x } } -\sqrt { x-1 }$
Hence

$I=\cfrac { 4 }{ \pi } \left( x\sec ^{ -1 }{ \sqrt { x } } -\sqrt { x-1 } \right) -x+c$

$\displaystyle \int xe^{-5x^2} \: sin \: 4x^2 dx = Ke^{-5x^2} (A \: sin \: 4x^2 + B \: cos \: 4x^2) + C$ . Then

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$\dfrac {1}{82}, 5, 4$
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$\dfrac {1}{82}, -5, 4$
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$\dfrac {1}{82}, -5, -4$
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$\dfrac {1}{82}, \dfrac{-1}{5}, \dfrac{-1}{4}$

Explanation

let , $I=\int { x{ e }^{ -5{ x }^{ 2 } }\sin { (4{ x }^{ 2 })dx } }$

put $t={ x }^{ 2 }\quad \Rightarrow dt=2xdx$

$I=\frac { 1 }{ 2 } \int { { e }^{ -5t } } \sin { (4t)dt }$

let , $K=\int { { e }^{ -5t } } \sin { (4t)dt }$

using integration by parts,

$={ e }^{ -5t }\int { \sin { (4t)dt } } -\int { (-5{ e }^{ -5t }) } \int { \sin { (4t)dt } }$

$=-{ e }^{ -5t }\frac { \cos { (4t) } }{ 4 } -\frac { 5 }{ 4 } \int { { e }^{ -5t } } \cos { (4t) } dt$

using integration by parts again,

$=-{ e }^{ -5t }\frac { \cos { (4t) } }{ 4 } -\frac { 5 }{ 4 } \left[ { e }^{ -5t }\frac { \sin { (4t) } }{ 4 } +\frac { 5 }{ 4 } \int { { e }^{ -5t } } \sin { (4t)dt } \right]$

$K=-{ e }^{ -5t }\frac { \cos { (4t) } }{ 4 } -\frac { 5 }{ 16 } { e }^{ -5t }\sin { (4t) } -\frac { 25 }{ 16 } K$

$K=-\frac { 1 }{ 41 } { e }^{ -5t }\left( 4\cos { (4t)+5\sin { (4t) } } \right)$

$K=-\frac { 1 }{ 41 } { e }^{ -5t }\left( 4\cos { (4t)+5\sin { (4t) } } \right)$

substituting $t={ x }^{ 2 }\quad$

$I=\frac { 1 }{ 2 } K=-\frac { 1 }{ 82 } { e }^{ -5{ x }^{ 2 } }\left( 4\cos { (4{ x }^{ 2 })+5\sin { (4{ x }^{ 2 }) } } \right)$

Threrefore , $A=5$ $K=-\frac { 1 }{ 82 }$

$\int e^{x^4}\, (x\, +\, x^3\, +\, 2x^5)e^{x^2}dx$ is equal to

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$\displaystyle \frac{1}{2} xe^{x^2}\, \cdot\, e^{x^4}\, +\, c$
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$\displaystyle \frac{1}{2} x^2\, e^{x^4}\, +\, c$
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$\displaystyle \frac{1}{2} e^{x^2}\, \cdot\, e^{x^4}\, +\, c$
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$\displaystyle \frac{1}{2} x^2 e^{x^2}\, \cdot\, e^{x^4}\, +\, c$

Explanation

$f(x)=\int { { e }^{ { x }^{ 2 } }{ e }^{ { x }^{ 4 } }(x+{ x }^{ 3 }+2{ x }^{ 5 })dx } \\ f(x)=\int { { e }^{ { x }^{ 2 } }{ e }^{ { x }^{ 4 } }2{ x }^{ 5 }dx } +\int { { e }^{ { x }^{ 2 } }{ e }^{ { x }^{ 4 } }(x+{ x }^{ 3 })dx } \\ f(x)=\dfrac { 1 }{ 2 } \int { { e }^{ { x }^{ 2 } }{ e }^{ { x }^{ 4 } }4{ x }^{ 5 }dx } +\dfrac { 1 }{ 2 } \int { { e }^{ { x }^{ 2 } }{ e }^{ { x }^{ 4 } }(2x+2{ x }^{ 3 })dx } \\ f(x)=\dfrac { 1 }{ 2 } \int { ({ x }^{ 2 }{ e }^{ { x }^{ 2 } }4{ x }^{ 3 }{ e }^{ { x }^{ 4 } })+{ e }^{ { x }^{ 4 } }[{ e }^{ { x }^{ 2 } }2x+{ { x }^{ 2 }e }^{ { x }^{ 2 } }2x]\quad dx } \\ f(x)=\dfrac { 1 }{ 2 } \int { { d(x }^{ 2 }{ e }^{ { x }^{ 2 } }{ e }^{ { x }^{ 4 } }) } \\ f(x)=\dfrac { 1 }{ 2 } { x }^{ 2 }{ e }^{ { x }^{ 2 } }{ e }^{ { x }^{ 4 } }+c$

$\displaystyle I = \int \left ( \frac {1}{x^{1/2} - x^{1/4}} + \frac {\log (1 + x^{1/4})}{x^{1/2} + x^{1/4}} \right ) dx$ , then I equals

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$\displaystyle 2 \sqrt x + 4x^{1/4} + \log |x^{1/4} - 1| + C$
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$\displaystyle 2(x^{1/2} - 6x^{1/4} + 3) \log (x^{1/4} + 1) + C - (x^{1/2} - 10x^{1/4} + 7) + 2(\log (x^{1/4} + 1))^2$
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$\displaystyle \sqrt x - x^{1/4} + 6 \log (x^{1/4} + 1) - 3x^{3/4} + C$
0%
none of these

Explanation

$I=\int \left( \cfrac { 1 }{ x^{ 1/2 }-x^{ 1/4 } } +\cfrac { \log (1+x^{ 1/4 }) }{ x^{ 1/2 }+x^{ 1/4 } } \right) dx$
Put

$x={ t }^{ 4 }$

$I=\int { \left( \cfrac { 4{ t }^{ 3 }dt }{ { t }^{ 2 }-t } +\cfrac { \log { \left( 1+t \right) } }{ { t }^{ 2 }+1 } 4{ t }^{ 3 } \right) } dt$
Let

$I={ I }_{ 1 }+{ I }_{ 2 }$
Where

${ I }_{ 1 }=4\int { \cfrac { { t }^{ 2 } }{ t-1 } dt } =4\int { \left( t+1+\cfrac { 1 }{ t-1 } \right) dt } \\ =2{ t }^{ 2 }+4t+\log { \left| t-1 \right| } \\ =2\sqrt { x } +4{ x }^{ 3/4 }+\log { \left| { x }^{ 1/4 }-1 \right| }$
And

${ I }_{ 2 }=4\int { \cfrac { \log { \left( 1+t \right) } }{ { t }+1 } } { t }^{ 2 }dt\\ t+1=u\\ { I }_{ 2 }=4\int { \cfrac { \log { u } }{ u } { \left( u-1 \right) }^{ 2 }du } \\ =4\int { \left( u-2+\cfrac { 1 }{ u } \right) \log { u } du } \\ =\left( 2{ u }^{ 2 }-8u \right) \log { u } -\left( { u }^{ 2 }-8u \right) +2{ \left( \log { u } \right) }^{ 2 }\\ =2\left( { x }^{ 1/2 }-6{ x }^{ 1/4 }+3 \right) \log { \left( { x }^{ 1/4 }+1 \right) } -\left( { x }^{ 1/2 }-10{ x }^{ 1/4 }+7 \right) +2{ \left( \log { \left( { x }^{ 1/4 }+1 \right) } \right) }^{ 2 }$

$\displaystyle \int\, sin^2\, (lnx)\, dx$ is equal to

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$\displaystyle \frac{2x}{15} (5 + 2sin(2lnx) + cos(2lnx)) + c$
0%
$\displaystyle \frac{x}{5} (5 + 2sin(2lnx) -cos(2lnx)) + c$
0%
$\displaystyle \frac{x}{10} (5 -2sin(2lnx)- cos(2lnx)) + c$
0%
$\displaystyle \frac{x}{10} (5 -2sin(2lnx) + sin(2lnx)) + c$

Explanation

Let

$I=\int { \sin ^{ 2 }{ \left( \ln { x } \right) } dx }$
Substitute

$\displaystyle t=\ln { x } \Rightarrow dt=\frac { dx }{ x }$

$\displaystyle \therefore I=\int { { e }^{ t }\sin ^{ 2 }{ t } dt } =\frac { 1 }{ 2 } \int { { e }^{ t }\left( 1-\cos { 2t } \right) } dt$

$\Rightarrow 2I=\int { { e }^{ t }dt-\int { { e }^{ t } } } \cos { 2t } dt={ e }^{ t }-{ I }_{ 1 }$

where

$\displaystyle { I }_{ 1 }=\int { { e }^{ t }\cos { 2t } } dt=\frac { { e }^{ t } }{ 5 } \left[ \cos { 2t } +2\sin { 2t } \right] +c$ ...............

$\displaystyle \because \int { { e }^{ ax } } \cos { bx } dx=\frac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \left[ a\cos { bx } +b\sin { bx } \right]$

$\displaystyle \therefore I=\frac { 1 }{ 10 } { e }^{ t }\left[ 5-2\sin { 2t } -\cos { 2t } \right]$

$\displaystyle =\frac { x }{ 10 } \left( 5-2\sin { \left( 2\ln { x } \right) - } \cos { \left( 2\ln { x } \right) } \right)$

$\displaystyle\int\, \cos\, 2\theta.\,\ln\, \displaystyle \frac {\cos\theta\, +\, \sin\, \theta}{\cos\, \theta\, -\, \sin\, \theta}\, d\theta$

0%
$\, \displaystyle \frac {1}{2}\, ln\, \left ( \displaystyle \frac {\cos\, \theta\, -\, \sin\, \theta}{\cos\, \theta\, +\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\sec\, 2\,\theta)\, +\, C$
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$\, \displaystyle \frac {1}{2}\, \ln\, \left ( \displaystyle \frac {\cos\, \theta\, -\, \sin\, \theta}{\cos\, \theta\, +\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\text{cosec}\, 2\,\theta)\, +\, C$
0%
$\, \displaystyle \frac {1}{2}\, \ln\, \left ( \displaystyle \frac {\cos\, \theta\, +\, \sin\, \theta}{\cos\, \theta\, -\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\text{cosec}\, 2\,\theta)\, +\, C$
0%
$\, \displaystyle \frac {1}{2}\, \ln\, \left ( \displaystyle \frac {\cos\, \theta\, +\, \sin\, \theta}{\cos\, \theta\, -\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\sec\, 2\,\theta)\, +\, C$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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