If $$\int f(x)\ dx $$ $$=\Psi(x)$$ , then $$\int x^{5}f(x^{3})\ dx $$ is equal to

$$\dfrac{1}{3}x^{3}\Psi(x^{3})-3\int x^{3}\Psi(x^{3})dx+C$$

$$\dfrac{1}{3}[x^{3}\Psi(x^{3})-\int x^{3}\Psi(x^{3}) dx ]+C$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 1

$\displaystyle \int_{0}^{1}\tan ^{-1}x\:dx.$

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$\displaystyle \frac{\pi }{2}-\frac{1}{2}\log 2.$
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$\displaystyle \frac{\pi }{4}+\frac{1}{2}\log 2.$
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$\displaystyle \frac{\pi }{2}-\frac{1}{4}\log 2.$
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$\displaystyle \frac{\pi }{4}-\frac{1}{2}\log 2.$

Explanation

$\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } } dx=\left[ x\tan ^{ -1 }{ x } \right] _{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ \frac { x }{ { x }^{ 2 }+1 } } dx$

$\displaystyle =\frac { \pi }{ 4 } -\left[ \frac { 1 }{ 2 } \log { \left( { x }^{ 2 }+1 \right) } \right] _{ 0 }^{ 1 }=\frac { \pi }{ 4 } -\frac { 1 }{ 2 } \log { 2 }$

$\int \dfrac {dx}{\sqrt {x^{10} - x^{2}}}; x > 1=$ ______

$+ C$ .

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$\dfrac {1}{4}\log |\sqrt {x^{10} - x^{2}} + x^{2}|$
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$\dfrac {1}{2}\log |x^{10} - x^{2}|$
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$-\dfrac {1}{4}\sec^{-1} (x^{4})$
0%
$\dfrac {1}{4}\sec^{-1} (x^{4})$

Explanation

Let

$I=\displaystyle \int \dfrac {dx}{\sqrt {x^{10} - x^{2}}}$

$=\displaystyle \int \dfrac {4x^{3}dx}{4x^{3} \times x \sqrt {x^{8} - 1}}$

$=\displaystyle \int \dfrac {4x^{3} dx}{4x^{4}\sqrt {x^{8} - 1}}$
Put

$x^{4} = t$

$\Rightarrow 4x^{3}dx = dt$

$\displaystyle \int \dfrac {dt}{4t \sqrt {t^{2} - 1}} = \dfrac {1}{4}\sec^{-1} (x^{4}) + C$ .

The value of

$\int log x dx$ is

0%
$\displaystyle x log \left ( \frac{x}{e} \right ) + c$
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$x^2 log x + c$
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$e^x log x + c$
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$(x + 1) log x + c$

Explanation

$\int log x. 1. dx = log x. \int dx - \int \displaystyle \left ( \frac{1}{x} .\int dx \right ) dx$
[Integration by parts]

$(log x) x - \int \displaystyle \left( \frac{1}{x} \right ) x. dx = x. log_e x - x + C$

$x. [log_e x - log_e e] + c = x. log_e \displaystyle \left ( \frac{x}{e} \right ) + c$

$\displaystyle \int x.e^{x}dx=$

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$x.e^{x}+c$
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$e^{x}(x-1)+c$
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$e^{x}(x+1)+c$
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$e^{x}(x+2)+c$

Explanation

$\int x\ e^{x}\ dx=x\int e^{x}\ dx-\int\ (e^{x}\ dx)dx$

$=xe^{x}-e^{x}+c$

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

$\int xe^{x}dx=\ e^{x}(x-1)+c$

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

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$e^{x}(x^{3}-3x^{2}+6x+6)+c$
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$e^{x}(x^{3}- 3x^{2}+6x-6)+c$
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$e^{x}(x^{3}+3x^{2}+6x+6)+c$
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$e^{x}(x^{3}- 3x^{2}-6x+6)+c$

Explanation

$\int x^{3}-e^{x}dx$
=

$x^{3}\int e^{x}dx-\int 3x^{2}-e^{x}dx$
=

$x^{3}e^{x}-3\int x^{2}e^{x}dx$

$\int x^{2}e^{x}=x^{2}e^{x}-2\int xe^{x}-2\int xe^{x}dx$

$\int xe^{x}=xe^{x}-\int e^{x}dx$
=

$x^{3}e^{x}-3[x^{2}e^{x}-2[xe^{x}-e^{x}]]$

$x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+c$

$\int x^{3}e^{x}dx=e^{x}(x^{3}-3x^{2}+6x-6)+c$

$\displaystyle \int\log\sqrt{x+1}dx=$

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$\displaystyle \frac{1}{2}[(x+1)\log (x+1)-x]+c$
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$\displaystyle \frac{1}{2}[(x-1)\log(x+1)-x]+c$
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$\displaystyle \frac{1}{2}[x \displaystyle \log(x+1)-\frac{x^{2}}{2}]+c$
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$\displaystyle \frac{1}{2}[(x+1)\displaystyle \log(x+1)-\frac{x}{2}]+c$

Explanation

Integrating by parts,

$\int 1 log\sqrt {x+1}\cdot dx$

$=log \sqrt {x+1}\int 1dx - \int \dfrac {d log \sqrt {x+1}}{dx}\cdot x\cdot dx.$

$x(log \sqrt{x+1}-\dfrac {1}{2}\int (\dfrac {1}{1+x.})\cdot x dx.$

$\dfrac {1}{2}log (x+1)^x-\dfrac {1}{2}[\int 1\cdot dx-log (1+x)]$

$=\dfrac {1}{2}x log (x+1)-\dfrac {1}{2}\cdot x+\dfrac {1}{2}log (1+x)$

$=\dfrac {1}{2}[log (x+1)[x+1]-x]+c\cdot$

$\displaystyle \int[\sin(\log x)+\cos(\log x)]dx=$

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$\displaystyle e^{x} \sin (\log x)+c$
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$\displaystyle e^{x} \cos (\log x)+c$
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$\displaystyle x\sin (\log x)+c$
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$\displaystyle x\cos (\log x)+c$

Explanation

$\int [sin (log x)+cos (log x)]dx$

$=\int sin (log x)dx+\int cos (log x)dx$

Integrating by parts,

$=x sin (log x)-\int cos (log x)dx+\int cos (log x)dx$

$=x sin (log x)+c$

The integral

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

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$(x-1)e^{x+ \frac{1}{x}} +c$
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$xe^{x+\frac{1}{x}} +c$
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$(x+1)e^{x+\frac{1}{x}} +c$
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$-xe^{x+\frac{1}{x}} +c$

Explanation

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

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

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

Using integration by parts

$=xe^{(x+\frac{1}{x})}+c$

The integral

$\displaystyle \int x { \cos^{ -1 }\left(\displaystyle \frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) dx }$ is equal to :
(Note :

$(x>0)$ )

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$-x + (1+x^2) \cot^{-1} x+c$
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$-x+(1+x^{2})\tan^{-1}x+c$
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$-x-(1+x^{2})\tan^{-1}xc$
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$x-(1+x^2)\cot^{-1}x+c$

Explanation

$\displaystyle \int x \cos ^{-1} \left ( \frac{1-x^2}{1+x^2} \right ) dx (x > 0)$

$x=\tan \theta dx = \sec ^2 \theta d \theta$

$I=\displaystyle \int \tan \theta \cos ^{-1}(\cos 2 \theta) \sec ^2 \theta d \theta$

$I=\displaystyle \int 2 \theta \tan \theta \sec ^2 \theta d \theta$

Using by parts

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

$I=2 \theta \displaystyle \int \tan \theta \sec ^2 \theta d \theta - \int 2 \cdot \int \tan \theta \sec ^2 \theta d \theta$

$=\displaystyle 2 \theta \frac{\tan ^2 \theta}{2} - 2 \times \frac{1}{2} \int \tan ^2 \theta d \theta$

$= \theta \tan ^2 \theta - [\int (-1+\sec ^2 \theta)d\theta]$ ................. $\sec^2 x=1+\tan^2 x$

$=\theta \tan ^2 \theta + \theta - \tan \theta +c$

$=\theta (1+ \tan ^2 \theta ) - \tan \theta +c$

Replacing $\theta$ by $x$ ,

$I = (\tan ^{-1}x ) ( 1+x^2) - x +c$

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

$\int f(x)\ dx$

$=\Psi(x)$ , then

$\int x^{5}f(x^{3})\ dx$ is equal to

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$\dfrac{1}{3}x^{3}\Psi(x^{3})-3\int x^{3}\Psi(x^{3})dx+C$
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$\dfrac{1}{3}x^{3}\Psi(x^{3})-\int x^{2}\Psi(x^{3})dx+C$
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$\dfrac{1}{3}[x^{3}\Psi(x^{3})-\int x^{3}\Psi(x^{3}) dx ]+C$
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$\dfrac{1}{3}[x^{3}\Psi(x^{3})-\int x^{2}\Psi(x^{3}) dx ]+C$

Explanation

$\int f(x) dx =\Psi(x)$

Let

$x^{3}=t$

$\Rightarrow 3x^{2}dx= dt$

Then

$\displaystyle \int x^{5}f(x^{3}) dx$

$=\dfrac{1}{3}\int tf (t) dt$ ....................(replacing the value of

$x^3$ &

$dx$ )

$=\dfrac{1}{3}[t\int f(t)dt-\int\{1\cdot\int f(t)\ dt\ \}\ dt\ ]$

$=\dfrac{1}{3}[t\Psi(t)-\int \Psi(t)dt+C].$

$=\dfrac{1}{3}[x^{3}\Psi(x^{3})-\int x^{2}\Psi(x^{3})dx]+C.$

$f''(x)=-f(x)$ and

$g(x)=f'(x)$ and

$\displaystyle{F}({x})=\left ({f}\left (\frac{{x}}{2}\right)\right)^{2}+\left ({g}\left (\frac{{x}}{2}\right)\right)^{2}$ and given that

$F(5)=5$ , then

$F(10)$ is equal to

0%
$5$
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$10$
0%
$0$
0%
$15$

Explanation

$\mathrm{f}''(\mathrm{x})=-\mathrm{f}(\mathrm{x})$ and

$\mathrm{f'}(\mathrm{x})=\mathrm{g}(\mathrm{x})$

$\Rightarrow \mathrm{f}''(\mathrm{x}).\ \mathrm{f'}(\mathrm{x})+\mathrm{f}(\mathrm{x}).\ \mathrm{f'}(\mathrm{x})=0$

$Integrating,$

$(\mathrm{f'}(\mathrm{x}))^{2}+(\mathrm{f}(\mathrm{x}))^{2}=\mathrm{c}\Rightarrow(\mathrm{f}(\mathrm{x})^{2}+(\mathrm{g}(\mathrm{x}))^{2}=\mathrm{c}$

$\Rightarrow \mathrm{F}(\mathrm{x})=\mathrm{c}$

$\Rightarrow \mathrm{F}(10)=5$

$\int \frac{ f(x) g'(x)-f'(x)g(x)}{f(x)g(x)}\begin{bmatrix} Log (g(x))-Log (f(x)) \end{bmatrix}dx=$

0%
$Log\begin{pmatrix} \frac {g(x)}{f(x)}\end{pmatrix}+C$
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$\frac {1 }{2 } \begin {vmatrix} Log \begin{pmatrix} \frac {g(x)}{f(x)}\end{pmatrix} \end {vmatrix}^2+C$
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$\dfrac {(g)x}{f(x)}Log \begin {pmatrix}\frac {g(x)}{f(x)}\end {pmatrix} +C$
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$Log\begin{bmatrix} \frac {g(x)}{f(x)} \end{bmatrix}-\frac {g(x)}{f(x)}+C$

Explanation

Put

$log(g(x_1))-log(f(x_1))=1$

$\int x log x dx$ is equal to

0%
$\displaystyle\frac{x^2}{4}(2log x-1)+c$
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$\displaystyle\frac{x^2}{2}(2log x-1)+c$
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$\displaystyle\frac{x^2}{4}(2log x+1)+c$
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$\displaystyle\frac{x^2}{2}(2log x+1)$

Explanation

Using ILATE

$\int \underset{II}{x}\underset{I}{log}x dx=logx\cdot \displaystyle\frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}dx$

$=\displaystyle\frac{x^2}{2}log x-\frac{1}{2}\frac{x^2}{2}+c$

$=\displaystyle\frac{x^2}{4}(2log x-1)+c$

The value of

$\int { { e }^{ \tan { \theta } } } \left( \sec { \theta } -\sin { \theta } \right) d\theta$ is equal to ?

0%
$-{ e }^{ \tan { \theta } }\sin { \theta } +C$
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${ e }^{ \tan { \theta } }\sin { \theta } +C$
0%
${ e }^{ \tan { \theta } }\sec { \theta } +C$
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${ e }^{ \tan { \theta } }\cos { \theta } +C$

Explanation

$\displaystyle I = \int e^{\tan x}. \sec x dx - \int e^{\tan x} .\sin x dx$

$\displaystyle I = \int e^{\tan \theta}. \sec\theta. d\theta - \int e^{\tan\theta}. \sin\theta . d\theta$

by using integration by parts.

$\displaystyle I = \int e^{\tan\theta}.\sec\theta \, d\theta + e^{\tan\theta}. \cos\theta - \int e^{\tan\theta}.(\sec^2\theta).\cos\theta.d\theta$

$\displaystyle I = \int e^{\tan\theta}.\sec\theta d\theta - \int e^{\tan\theta}\sec \, \theta.d\theta +e^{\tan\theta}.\cos\theta$

$\displaystyle I = e^{\tan\theta}. \cos\theta + c.$

$\displaystyle \int { \log { \left( \log { x } \right) +\dfrac { 1 }{ \log { x } } } } dx$

0%
$x\log { \left( \log { x } \right) +c }$
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$x\log { x } +c$
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$x+c$
0%
$None$

Explanation

$\int {\left[ {\log x\left( {\log x} \right) + \dfrac{1}{{\log x}}} \right]} dx$

$= \int {1.\log \left( {\log x} \right)dx + \int {\dfrac{1}{{\log x}}} }$

$= \log \left( {\log x} \right).x-\int {\dfrac{1}{{\log x}} \times \dfrac{1}{x} \times xdx + \int {\dfrac{{dx}}{{\log x}}} }$

$= x\log \left( {\log x} \right) - \int {\dfrac{{dx}}{{\log x}} + \int {\dfrac{{dx}}{{\log x}}} }$

$= x\log \left( {\log x} \right) + c$

The integral of

$\displaystyle\int e^{x} (\sin x+\cos x)dx$ is

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$e^{ x} \cos x + c$
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$e^{x}\sin x + c$
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$e^{ x}\sec x + c$
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none of this

Explanation

Now,

$\displaystyle\int e^{x} (\sin x+\cos x)dx$

$=\displaystyle\int e^{x} \sin x \ dx$

$+\displaystyle\int e^{x} \cos x\ dx$

$=e^x(\sin x)-\displaystyle\int (\cos x).e^x \ dx$

$+\displaystyle\int e^{x} \cos x\ dx$

$=e^x(\sin x)+c$ [ Where

$c$ is integrating constant]

$\displaystyle\int { \frac { { 2 }^{ x } }{ \sqrt { 1-{ 4 }^{ x } } } } dx=K\sin ^{ -1 }{ \left( { 2 }^{ x } \right) } +C$ , then the value of

$K$ is equal to

0%
$\ell n\ 2$
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$\dfrac {1}{2} \ell n\ 2$
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$\dfrac {1}{2}$
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$\dfrac {1}{\ell n\ 2}$

Explanation

$\displaystyle\int \dfrac{2^{x}}{\sqrt{1-4^{x}}}dx$

put

$t=2^{x}\implies dt=2^{x}\ln 2 dx\implies dx=\dfrac{dt}{t\ln 2}$

$\displaystyle\int \dfrac{1}{\sqrt{1-t^{2}}}\times \dfrac{dt}{\ln 2}=\dfrac{1}{\ln 2}\text{sin}^{-1}(2)^{x}+C\implies K=\dfrac{1}{\ln 2}$

The value of integral

$\int {{{\tan }^{ - 1}}\left( {\dfrac{{{x^3}}}{{1 + {x^2}}}} \right)} + {\tan ^{ - 1}}\left({\dfrac{{1 + {x^2}}}{{{x^3}}}} \right)dx$ is equal to

0%
$1$
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$- \dfrac{\pi }{2} + c$
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$\dfrac{\pi }{2} + c$
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$\left( {\dfrac{\pi }{2}} \right)x + c$

Explanation

$\int {{{\tan }^{ - 1}}\left( {{{{x^3}} \over {1 + {x^2}}}} \right) + } {\tan ^{ - 1}}\left( {{{1 + {x^2}} \over {{x^3}}}} \right)dx$

$= \int {{{\tan }^{ - 1}}\left[ {{{{{{x^3}} \over {1 + {x^2}}} + {{1 + {x^2}} \over {{x^3}}}} \over {1 - \left( {{{{x^3}} \over {1 + {x^2}}}} \right)\left( {{{1 + {x^2}} \over {{x^3}}}} \right)}}} \right]} dx$

$= \int {{{\tan }^{ - 1}}\left( {{{{{{x^6} + {{\left( {1 + {x^2}} \right)}^2}} \over {\left( {1 + {x^2}} \right)\left( {{x^3}} \right)}}} \over 0}} \right)} dx$

$= \int {{\pi \over 2}dx}$

$= {\pi \over 2}x + c$

$\int ( e^{\log x} + \sin x) \cos x \ dx$ is equal to

0%
$x \sin x + \cos x - \dfrac {\sin^2x}{2}+ c$
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$x \cos x - \sin^2x+ c$
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$x \sin x + \cos x - (\cos^2x)/2+ c$
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$x^2 \sin^x + \cos x - \sin^3 x + c$

Explanation

$\int (e^{\log x}+\sin x)\cos x \ dx$

$=\int (x+\sin x)\cos x \ dx$

$=\int x \cos x \ dx+\int \sin x \cos x \ dx$

$=x \sin x-\int \sin x \ dx+\int \sin x \ d \sin x$

$=x\sin x+\cos x+\dfrac{\sin^{2}x}{2}+c$

$\int_{}^{} {x{{\sin }^{ - 1}}xdx}$

0%
$\frac{1}{4}{\sin ^{ - 1}}x\left( {2x + 1} \right) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
0%
$\frac{1}{4}{\sin ^{ - 1}}x\left( {2{x^2} + 1} \right) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c]$
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$\frac{1}{4}{\cos ^{ - 1}}x\left( {2x + 1} \right) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
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$\frac{1}{4}{\sin ^{ - 1}}x\left( {2x + 1} \right) - \frac{{x\sqrt {1 - {x^2}} }}{4} + c$

Explanation

$\begin{array}{c}I = x{\sin ^{ - 1}}xdx\\ = \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} - \int_{}^{} {\frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }}dx} \\{I_1} = \int_{}^{} {\frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }}} dx = \frac{1}{2}\left[ {\int_{}^{} {\frac{{{x^2} - 1}}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - {x^2}} }}} } \right]dx\\ = \frac{1}{2}\left[ {\int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}dx - \int_{}^{} {\sqrt {1 - {x^2}} } dx} } \right]\end{array}$
Now,

$\begin{array}{l}\int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x} \\\int_{}^{} {\sqrt {1 - {x^2}} dx = \frac{{{{\sin }^{ - 1}}x}}{2}} + \frac{{x\sqrt {1 - {x^2}} }}{2}\\I = \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} - \frac{{{{\sin }^{ - 1}}x}}{4} + \frac{{x\sqrt {1 - {x^2}} }}{4}\end{array}$

Solve

$\int {\dfrac{1}{{\sqrt[{}]{{9 - 25{x^2}}}}}} dx$

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$\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C$
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${{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C$
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$\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{3x}{5} \right)+C$
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${{\sin }^{-1}}\left( \dfrac{3x}{5} \right)+C$

Explanation

We have,

$I=\int{\dfrac{dx}{\sqrt{9-25{{x}^{2}}}}}$

$I=\int{\dfrac{dx}{5\sqrt{\dfrac{9}{25}-{{x}^{2}}}}}$

$I=\dfrac{1}{5}\int{\dfrac{dx}{\sqrt{{{\left( \dfrac{3}{5} \right)}^{2}}-{{x}^{2}}}}}$

We know that

$\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$

Therefore,

$I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{x}{\dfrac{3}{5}} \right)+C$

$I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C$

Hence, this is the correct answer.

Find the general solution of

$\frac{{dy}}{{dx}} = \frac{{2y}}{x}$

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$y=e^{5\log x +c}$
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$y=e^{3\log x +c}$
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$y=e^{2\log x +c}$
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None of these

Explanation

$\dfrac{dy}{dx}=\dfrac{2y}{x}$

$\rightarrow\dfrac{dy}{y}=\dfrac{2dx}{x}$

integrating both sides,

$\int \dfrac{dy}{y}=\int \dfrac{2dx}{x}$

$\rightarrow\log y=2\log x +c$

$\rightarrow y=e^{2\log x +c}$

The value of

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

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$\cfrac { -\log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$
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$\cfrac { \log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$
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$\cfrac { \log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } + C$
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$\cfrac { -\log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } +C$

Explanation

$\int \dfrac{lnx}{(x+1)^{2}}$

$lnx\int \dfrac{dx}{(x+1)^{2}}-\int \dfrac{1}{x}\int \dfrac{dx}{(x+1)^{2}}$

$lnx\dfrac{(x+1)^{-2+1}}{-2+1}-\int \dfrac{1}{x}\dfrac{(x+1)^{-2+1}}{-2+1}dx$

$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}\dfrac{1}{(x+1)}dx$

$-\dfrac{lnx}{(x+1)}+\int \dfrac{(x+1)-(x)}{x(x+1)}$

$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}-\dfrac{1}{1+x}dx$

$-\dfrac{lnx}{(1+x)}+lnx-ln(1+x)+C$

$\displaystyle \int \sin x \log (\sec x+\tan x)dx=f(x)+x+c$ then

$f(x)=$

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$\cos x \log(\sec x + \tan x)+c$
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$\sin x\log(\sec x+\tan x)+c$
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$- \cos x \log\sec x+\tan x)+c$
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$-\cos x\log \sec x+c$

Explanation

$\displaystyle \int\ sinx\ log\ (secx+\tan x)dx$

$\displaystyle =\ log\ (secx+\tan x)\int sinx\ dx$

$\displaystyle -\int\left ( \frac{d}{dx}(log\ (secx+\tan x))\int\ sinx\ dx \right )dx$

$\displaystyle =-log(secx+\tan x)\ cosx\ +\ \int\ secx\ cosx\ dx$

$\displaystyle =-\ cosx\ log\ (secx+\tan x )+x+c$
So,

$\ f(x)=-cosx\ log\ (secx+\tan x)$

$\int [\frac{f(x)g'(x)-f'(x)g(x)}{f(x)g(x)}].[log (g(x))-log(f(x))]dx=$

0%
$log\frac{g(x)}{f(x)}+c$
0%
$\frac{1}{2}(log\frac{g(x)}{f(x)})^{2}+c$
0%
$\frac{g(x)}{f(x)}log(\frac{g(x)}{f(x)})+c$
0%
$\frac{f(x)}{g(x)}log(\frac{g(x)}{f(x)})+c$

Explanation

$\displaystyle\left[\dfrac{f(x)g'(x)-f'(x)g(x)}{f(x)g(x)}\right][log g(x)-log f(x)]dx=I$

$I=\displaystyle\int \dfrac{f(x)g'(x)-f'(x)g(x)}{f^2(x)}\cdot \dfrac{f(x)}{g(x)}\cdot log\dfrac{g(x)}{f(x)}dx$

$I=\displaystyle\int \left[\dfrac{f(x)g'(x)-f'(x)g(x)}{f^2(x)}\right]\dfrac{log\dfrac{g(x)}{f(x)}}{g(x)/f(x)}dx$

Let

$log \dfrac{g(x)}{f(x)}=u$

$\dfrac{1}{\dfrac{g(x)}{f(x)}}\left[\dfrac{f(x)g'(x)-f'(x)g(x)}{f^2(x)}\right]dx=du$

$I=\displaystyle\int u du$

$I=\dfrac{u^2}{2}+c$

$I=\dfrac{1}{2}\left(log\dfrac{g(x)}{f(x)}\right)^2+c$ .

1179232_1241988_ans_f94f95409ddb455c84320c56d9bb33b0.jpg

$\frac{1}{2}\int \sqrt{x^{3}-8x^{2}dx}$ equals

0%
${\frac{(x-8)^{5/2}}{5}+\frac{8(x-8)^{3/2}}{3}}+c$
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${\frac{(x-8)^{5/2}}{5}+\frac{8}{3}(x-8)^{3/2}}+c$
0%
${\frac{(x-8)^{5/2}}{3}+\frac{8}{5}(x-8)^{3/2}}+c$
0%
None of these

Explanation

$I=\dfrac{1}{2}\displaystyle\int \sqrt{x^3-8x^2}dx$

$=\dfrac{1}{2}\displaystyle\int x\sqrt{x-8}dx$

Using By-parts for x &

$\sqrt{x-8}$

$=\dfrac{1}{2}(x\displaystyle\int \sqrt{x-8}dx-\displaystyle\int \dfrac{d(x)}{dx}(\displaystyle\int \sqrt{x-8}dx)dx$

$=\dfrac{1}{2}(x\dfrac{(x-8)^{3/2}}{3/2}-\displaystyle\int \dfrac{(x-8)^{3/2}}{3/2}dx)$

$I=\dfrac{1}{2}(\dfrac{x(x-8)^{3/2}}{3/2}-\dfrac{(x-8)^{5/2}}{3/2\times 5/2}+c)$

$I=\dfrac{x(x-8)^{3/2}}{3}-\dfrac{2}{15}(x-8)^{5/2}+c$

Answer is option D

1179257_1245402_ans_a862fd1ff6764589ab0b17489f614893.jpg

$\int\ \sec\theta\ d\theta$

0%
$\dfrac{(\sec\theta+\tan\theta)}{2}+c$
0%
$\dfrac{(\sec\theta+\tan\theta)}{3}[2+4\tan\theta(\sec\theta+\tan\theta)]+c$
0%
$\ln {(|\sec\theta+\tan\theta|)}+c$
0%
$\dfrac{3(\sec\theta+\tan\theta)}{2}[2+\tan\theta(\sec\theta+\tan\theta)]+c$

$\int {\dfrac{{\cot \sqrt x }}{{2\sqrt x }}dx}$ is equal to

$= \_\_\_\_\_ + C.$

0%
$2\log |\sin \sqrt x |$
0%
$\log |\sin \sqrt x |$
0%
$\dfrac{1}{2}\log |\sin \sqrt x |$
0%
None of these

Explanation

$I=\int { \dfrac { { \cot \sqrt { x } } }{ { 2\sqrt { x } } } dx }$

Let

$t=\sqrt { x }$

$\dfrac { 1 }{ { 2\sqrt { x } } } dx=dt$

Therefore,

$I=\int { \cot dt }$

$I=\log \sin t+C$

On putting the value of

$t$ , we get

$I=\log \sin \sqrt x+C$

Hence, this is the answer.

Evaluate :

$\int \cos^3 x e^{\log \sin x} dx$

0%
$-\dfrac{\cos ^4x}{4}+C$
0%
$\dfrac{\sin x}{x^2}+C$
0%
$-\dfrac{\sin^3x}{3}+C$
0%
None of these

Explanation

$I=\int \cos^3 x \ e^{\log \sin x} dx$

$I=\int \cos ^3 x \sin x \ dx$

Putting

$\cos x =t$ and

$-\sin x \ dx=dt$

or,

$\sin x \ dx=-dt$ , we get,

$I=-\int t^3 \ dt=-\dfrac{t^4}{4}+C=-\dfrac{\cos ^4x}{4}+C$

The value of

$\int { (x-1) } { e }^{ -x }$ dx is equal to

0%
$-{ xe }^{ x }+C$
0%
${ xe }^{ x }+C$
0%
$-{ xe }^{ -x }+C$
0%
${ xe }^{ -x }+C$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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