If $$I=\int e^x\frac{(1-x)^2}{(1+x^2)^2}dx $$ , then I =

$$ e^x\frac{1-x^2}{(1+x^2)}+C$$

$$ e^x\frac{1-2x}{(1+x^2)^2}+C$$

If $$l_{n}=\displaystyle \int{\dfrac{t^{n}}{1+t^{2}}}dt$$ then

$$l_{n+2}=\dfrac{t^{n}}{n}-nl_{n}$$

$$l_{n+1}=\dfrac{t^{n+1}}{n+1}l_{n}$$

$$l_{n+1}=\dfrac{t^{n-1}}{n-1}l_{n}$$

$$l_{n21}=\dfrac{t^{n+1}}{n+1}l_{n}$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 7

Evaluate

$\displaystyle \int_{}^{} {x\sqrt {\frac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} dx }$

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$\displaystyle \frac{1}{2}{a^2}{\cos ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \frac{1}{2}\sqrt {{a^4} + {x^4}} + C$
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$\displaystyle \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \sqrt {{a^4} + {x^4}} + C$
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$\displaystyle \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \frac{1}{2}\sqrt {{a^4} - {x^4}} + C$
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$\displaystyle \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{{x^2}}}{{{a^2}}}} \right) + \frac{1}{2}\sqrt {{a^4} - {x^4}} + C$

Explanation

Consider

$i=\displaystyle \int{x\dfrac{\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}}}dx$

put

$x^2=t$

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

$I=\dfrac{1}{2}\displaystyle \int{\dfrac{\sqrt{a^2-t}}{\sqrt{a^2+t}}}dt=\dfrac{1}{2}\displaystyle \int{\dfrac{a^2-t}{\sqrt{a^4-t^2}}}dt$

$I=\dfrac{1}{2}\displaystyle \int{\dfrac{a^2dt}{\sqrt{a^4-t^2}}}-\dfrac{1}{2}\displaystyle \int{\dfrac{t}{\sqrt{a^4-t^2}}}dt$

$=\dfrac{a^2}{2}\sin^{-1}\dfrac{t}{a^2}+\dfrac{1}{4}\displaystyle \int{\dfrac{-2tdt}{\sqrt{a^4-t^2}}}$ put

$a^4-t^2=u\\-2tdt=du$

$=\dfrac{a^2}{2}\sin^{-1}\dfrac{x^2}{a^2}+\boxed{\dfrac{1}{4}\displaystyle \int{\dfrac{du}{\sqrt u}}}\rightarrow \dfrac{1}{2}\sqrt u+c$

$=\dfrac{a^2}{2}\sin^{-1}\dfrac{x^2}{a^2}+\dfrac{1}{2}\sqrt{a^4-x^4}+c$ .

Solve:

$\int 4x^2 e^{2x^3} \ dx$

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$\dfrac{2}{3}e^{2x^3}+C$
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$\dfrac{4}{3}e^{2x^3}+C$
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$\dfrac{3}{2}e^{2x^3}+C$
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$\dfrac{1}{3}e^{2x^3}+C$

Explanation

We have,

$I=\int 4x^2 e^{2x^3} \ dx$

Let

$t=2x^3$

$\dfrac{dt}{dx}=6x^2$

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

Therefore,

$I=\dfrac{2}{3}\int e^{t} \ dt$

$I=\dfrac{2}{3}e^t+C$

On putting the value of

$t$ , we get

$I=\dfrac{2}{3}e^{2x^3}+C$

Hence, this is the answer.

The value of $\displaystyle\int {\dfrac{{\ln n\left( {1 - \left( {\dfrac{1}{x}} \right)} \right)dx}}{{x\left( {x - 1} \right)}}}$ is

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$\dfrac{1}{2}\left[ {m{{\left( {1 - \frac{1}{x}} \right)}^2}} \right] + C$
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$\dfrac{1}{2}{\left[ {m\left( {1 + \frac{1}{x}} \right)} \right]^2} + C$
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$\dfrac{1}{2}\ell n\left( {x\left( {x - 1} \right)} \right) + C$
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$\dfrac{1}{2}{\left[ {\ell n(x(x - 1))} \right]^2} + C$

Explanation

Consider, $\displaystyle I=\int {{{ln\left( {1 - {1 \over x}} \right)} \over {x\left( {x - 1} \right)}}} dx$

$\displaystyle I = \int {{{\ln\left( {1 - {1 \over x}} \right)} \over {{x^2}\left( {1 - {1 \over x}} \right)}}} dx$

Put $\ln\left( {1 - {\dfrac 1 x}} \right) = t$ $\Rightarrow$ $\displaystyle {1 \over {\left( {1 - {\dfrac 1 x}} \right)}} \times {\dfrac {1} {{x^2}}}dx = dt$

$\displaystyle I= \int {tdt}$

$\displaystyle I= {{{t^2}} \over 2} + c$

$\displaystyle I= {1 \over 2}{\left[ {ln\left( {1 - {1 \over x}} \right)} \right]^2} + c$

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

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$e^x\frac{1-x^2}{(1+x^2)}+C$
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$e^x\frac{1-x}{(1+x^2)}+C$
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$e^x\frac{1}{(1+x^2)}+C$
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$e^x\frac{1-2x}{(1+x^2)^2}+C$

Explanation

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

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

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

$-$

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

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

$-$

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

By integration by parts :

$\int [f(x) g(x)]dx=f(x)\cdot \int g(x)dx-\int[f'(x) \int g(x) dx]dx$

By integration by parts of first integral

$I=$

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

$-$

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

$I= e^x\frac{1}{(1+x^2)}$

$+$

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

$-$

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

$I= e^x\frac{1}{(1+x^2)}$

$+$

$C$

The value of

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

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$x{e^{{{\tan }^{ - 1}}x}} + C$
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${x^2}{e^{{{\tan }^{ - 1}}x}} + C$
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$\dfrac{1}{x}{e^{{{\tan }^{ - 1}}x}} + C$
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$x{e^{{{\cot }^{ - 1}}x}} + C$

Explanation

Consider the given equation.

$I=\int{{{e}^{{{\tan }^{-1}}x}}\left( \dfrac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)}dx$

$I=\int{\left( \dfrac{{{e}^{{{\tan }^{-1}}x}}+x{{e}^{{{\tan }^{-1}}x}}+{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}} \right)}dx$ ……… (1)

Let $t=x{{e}^{{{\tan }^{-1}}x}}$

$t=x{{e}^{{{\tan }^{-1}}x}}$

$\dfrac{dt}{dx}={{e}^{{{\tan }^{-1}}x}}\times 1+x\left( {{e}^{{{\tan }^{-1}}x}} \right)\times \dfrac{1}{1+{{x}^{2}}}$

$\dfrac{dt}{dx}={{e}^{{{\tan }^{-1}}x}}+\dfrac{x{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$

$dt=\left( \dfrac{{{e}^{{{\tan }^{-1}}x}}+{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}+x{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}} \right)dx$

From equation (1), we get

$I=\int{1dt}$

$I=t+C$

On putting the value of $t$ , we get

$I=x{{e}^{{{\tan }^{-1}}x}}+C$

Hence, this is the answer.

Evaluate

$\displaystyle\int{\dfrac{x}{\sqrt{{{x}^{2}}+2}}dx}$

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$I=\sqrt{{{x}^{2}}-2}+C$
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$I=\sqrt{{{x}^{2}}+2}+C$
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$I=\sqrt{{{x}^{3}}+2}+C$
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$I=\sqrt{{{x}^{3}}-2}+C$

Explanation

Consider the given integral.

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

Let $t={{x}^{2}}+2$

$\dfrac{dt}{dx}=2x+0$

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

Therefore,

$I=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}}$

$I=\dfrac{1}{2}\left( 2\sqrt{t} \right)+C$

$I=\sqrt{t}+C$

On putting the value of $t$ , we get

$I=\sqrt{{{x}^{2}}+2}+C$

Hence, this is the answer.

The function

$f(x)$ satisfying the equation

$f^{2}(x)+4f'(x).f(x)+[f'(x)]^{2}=0$ is-

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$f(x)=c.e^{(2-\sqrt {3}x)}$
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$f(x)=c.e^{-(2-\sqrt {3}x)}$
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$f(x)=c.e^{(\sqrt {3}-2)x}$
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$f(x)=c.e^{-(2+\sqrt {3})x}$

Explanation

Given,

$f^{2}(x)+4f'(x).f(x)+[f'(x)]^{2}=0$

or,

$f'(x)=\dfrac{-4\pm \sqrt{12}}{2} f(x)$

or,

$\dfrac{d\{f(x)\}}{f(x)}=(-2\pm \sqrt{3})dx$

Now integrating both sides we get,

$f(x)=ce^{(-2\pm\sqrt{3})x}$ . [ Where

$c$ is integrating constant]

The value of

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

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$- x{e^x} + C$
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$x{e^x} + C$
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$- x{e^{ - x}} + C$
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$x{e^{ - x}} + C$

Explanation

$\int (x-1) e^{-x} \ dx$ .

$-x = t - dx= dt$

$= \int (t+1) e^{t} \ dt$

$= (t+1 )e^{t} - \int e^{t} \ dt$ [Integration by parts].

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

$= te^{t} + c$

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

$C$ is correct.

$\int ( 1+3x+3x^2+4x^3+........)dx (|x| <1)$ -

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$(1+x)^{-1}+c$
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$(1-x)^{-1}+c$
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$(1+x)^{-2}+c$
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None of these

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

$K$ is equal to

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

Explanation

Consider the following integral.

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

Solution:

Let and differentiate w.r.t “x” we get.

$t={{2}^{x}}$

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

$\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx\,\,\,\,\,\,\,\,......\left( 1 \right)\$

Put the value $dx\,\,$ of in eq . (1) we get.

$=\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx\,\,\,\,\,\,\,\,......\left( 1 \right)$

$=\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{{{2}^{x}}\ln 2\sqrt{1-{{\left( {{t}^{2}} \right)}^{x}}}}}dx$

$=\int\limits_{{}}^{{}}{\dfrac{1}{\ln 2\sqrt{1-{{\left( {{t}^{2}} \right)}^{x}}}}}dx\,$

$=\dfrac{1}{\ln 2}{{\sin }^{-1}}x+C$

Hence, the value of $[K=\dfrac{1}{\ln 2}]$

Hence, this is the correct answer .

$\int { \cos { \left( \log { x } \right) } } dx=............\quad +\quad c$

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$\dfrac { x }{ 2 } \left[ \cos { \left( \log { x } \right) } +\sin { \left( \log { x } \right) } \right]$
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$\dfrac { x }{ 4 } \left[ \cos { \left( \log { x } \right) } +\sin { \left( \log { x } \right) } \right]$
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$\dfrac { x }{ 2 } \left[ \cos { \left( \log { x } \right) } -\sin { \left( \log { x } \right) } \right]$
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$\dfrac { x }{ 2 } \left[ \sin { \left( \log { x } \right) } +\cos { \left( \log { x } \right) } \right]$

Explanation

Consider the given integral.

$I=\int{\cos \left( \log x \right)dx}$

Let $t=\log x$

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

$xdt=dx$

Therefore,

$I=\int{{{e}^{t}}\cos tdt}$

$I=\cos t{{e}^{t}}-\int{\left( -\sin t \right){{e}^{t}}}dt$

$I={{e}^{t}}\cos t+\int{\left( \sin t \right){{e}^{t}}}dt$

$I={{e}^{t}}\cos t+\sin t{{e}^{t}}-\int{\cos t{{e}^{t}}}dt$

$I={{e}^{t}}\cos t+\sin t{{e}^{t}}-I$

$2I={{e}^{t}}\left( \cos t+\sin t \right)+C$

$I=\dfrac{{{e}^{t}}}{2}\left( \cos t+\sin t \right)+C$

On putting the value of $'t'$ , we get

$I=\dfrac{{{e}^{\log x}}}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C$

$I=\dfrac{x}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C$

Hence, this is the answer.

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

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$\dfrac {1}{2}[\sec^{2}x-\tan x]+c$
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$\dfrac {1}{2}[x\sec^{2}x-\tan x]+c$
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$\dfrac {1}{2}[x\sec^{2}x+\tan x]+c$
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$\dfrac {1}{2}[\sec^{2}x+\tan x]+c$

Explanation

Consider the following integral .

$I=\int xsinxse{{c}^{3}}xdx=\int\limits_{{}}^{{}}{x\tan x{{\sec }^{2}}x}$

let $t=\tan x$ and differentiate both side w.r.t x, we get.

$\dfrac{dt}{{{\sec }^{2}}x}=dx$

$=\int\limits_{{}}^{{}}{\dfrac{t{{\tan }^{-1}}t{{\sec }^{2}}x}{{{\sec }^{2}}x}}dt$

$=\int\limits_{{}}^{{}}{t{{\tan }^{-1}}t}dt$

$={{\tan }^{-1}}t\int\limits_{{}}^{{}}{tdt-\dfrac{1}{2}\int\limits_{{}}^{{}}{\dfrac{1}{1+{{t}^{2}}}}}{{t}^{2}}dt$

$={{\tan }^{-1}}t\int\limits_{{}}^{{}}{tdt-\dfrac{1}{2}\left( \int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}dt} \right)}$

$=\dfrac{{{\tan }^{2}}x}{2}{{\tan }^{-1}}\left( \tan x \right)+\dfrac{{{\tan }^{2}}x}{2}{{\tan }^{-1}}\left( \tan x \right)-\dfrac{\tan x}{2}+C$

$={{\tan }^{2}}x{{\tan }^{-1}}\left( \tan x \right)-\dfrac{\tan x}{2}+C$

Hence, this is the correct answer.

$\int { P\left( x \right) { e }^{ kx }dx=Q\left( x \right) { e }^{ 4x }+C }$ , where

$P(x)$ is polynomial of degree

$n$ and

$Q(x)$ is polynomial of degree

$7$ . Then the value of

$n+7+k+\lim _{ x\rightarrow \infty }{ \dfrac { P\left( x \right) }{ Q\left( x \right) } }$ is:

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$18$
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$19$
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$20$
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$22$

Evaluate using limit of sum:

$\displaystyle \int_{1}^{3} {(x+1)^2}dx$

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$26$
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$28$
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$30$
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$32$

Explanation

$\displaystyle \int_{1}^{3} {(x+1)^2}dx$

Let

$t=(x+1)\implies dt=dx$

$x \to 1\to 3$

$t\to 2 \to 4$

$\displaystyle \int _2^4 t^2 dt$

$\left.\dfrac {t^3}3\right]_2^4$

$\dfrac{4^3}{2}-\dfrac {2^3}{2}$

$32-4=28$

$\int _{ 0 }^{ \pi /4 }{ \cfrac { \sec ^{ 2 }{ x } }{ \left( 1+\tan { x } \right) \left( 2+\tan { x } \right) } dx }$ equals:

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$\log _{ e }{ \cfrac { 2 }{ 3 } }$
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$\log _{ e }{ 3 }$
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$\cfrac { 1 }{ 2 } \log _{ e }{ \cfrac { 4 }{ 3 } }$
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$\log _{ e }{ \cfrac { 4 }{ 3 } }$

Explanation

$I=\displaystyle \int_0^{\pi /4}\dfrac {\sec^2 x\ dx}{(1+\tan x)(2+\tan x)}$

Let

$\tan x=t$

$\Rightarrow \sec^2 x\ dx=dt$

$\therefore \ I=\displaystyle \int_0^{1}\dfrac {dt}{(1+t)(2+t)}=\displaystyle \int_0^{1}\dfrac {(2+t)-(1+t)}{(2+t)(1+t)}$

$=\displaystyle \int_0^{1}\dfrac {1}{1+t}-\dfrac {1}{2+t}$

$=[\ln (1+t)-\ln (2+t)]_0^1$

$=(\ln 2-\ln 1)-(\ln 3-\ln 2)$

$=2\ln 2-\ln 3$

$=\ln (4/3)\quad =(D)$

$l_{n}=\displaystyle \int{\dfrac{t^{n}}{1+t^{2}}}dt$ then

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$l_{n+2}=\dfrac{t^{n}}{n}-nl_{n}$
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$l_{n+1}=\dfrac{t^{n+1}}{n+1}l_{n}$
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$l_{n+1}=\dfrac{t^{n-1}}{n-1}l_{n}$
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$l_{n21}=\dfrac{t^{n+1}}{n+1}l_{n}$

$\int {{e^x}\left[ {{\mathop{\rm tanx}\nolimits} - log\left( {\cos x} \right)} \right]} dx =$

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${e^x}\log \left( {\sec x} \right) + c$
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${e^x}\log \left( {co\sec x} \right) + c$
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${e^x}\log \left( {\cos x} \right) + c$
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${e^x}\log \left( {\sin x} \right) + c$

Explanation

$\displaystyle\int e^x\left[\tan x-log (\cos x)\right]dx$

$=\displaystyle\int e^x\left[\tan x+log (\sec x)\right]d$

$=\displaystyle\int \dfrac{d}{dx}\left\{e^xlog(\sec x)\right\}dx$

$=e^xlog (\sec x)+c$ .

$\displaystyle \int \dfrac{dx}{\sqrt{\sin^3 x \cos^5 x}} = a \sqrt{\cot x } + b \sqrt {\tan^3x} + c$ where c is an arbitrary constant of integration then the values of

$'a'$ and

$'b'$ are respectively :

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$-2$ & $\dfrac{2}{3}$
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$2$ & $-\dfrac{2}{3}$
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$2$ & $\dfrac{2}{3}$
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None of these

Explanation

$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3x\cdot \cos^5x}}=a\sqrt{\cot x}+b\sqrt{\tan^3x}+c$ .......(1).

Now,

$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3x\cdot \cos^5x}}$

$=\displaystyle\int \dfrac{cosec^2 xdx}{\sqrt{\cot x.\cos^4 x}}$ [ Dividing the numerator and the denominator by

$\sin^2 x$ ]

$=\displaystyle\int \dfrac{cosec^2 x.\sec^2 xdx}{\sqrt{\cot x}}$

$=\displaystyle\int \dfrac{cosec^2 x.(1+\tan^2 x)dx}{\sqrt{\cot x}}$

$=\displaystyle\int \dfrac{cosec^2 x.dx}{\sqrt{\cot x}}$

$+\displaystyle\int \dfrac{cosec^2 x.\tan^2 x\ dx}{\sqrt{\cot x}}$

$=\displaystyle\int \dfrac{cosec^2 x.dx}{\sqrt{\cot x}}$

$+\displaystyle\int \dfrac{\sec^2 x\ dx}{\sqrt{\cot x}}$

$=-\displaystyle\int \dfrac{d(\cot x)}{\sqrt{\cot x}}$

$+\displaystyle\int {\sqrt{\tan x}d(\tan x)}$

$=-{2}\sqrt{\cot x}+\dfrac{2}{3}\sqrt{\tan^3x}+c$ .

Comparing this with (1) we get,

$a=-2$ and

$b=\dfrac{2}{3}$ .

$\displaystyle \int (x^2-x+5)\, dx$

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$\dfrac {x^3}3-\dfrac{x^2}2+5x+c$
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$\dfrac {x^3}3+\dfrac{x^2}2+5x+c$
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$\dfrac {x^2}2-\dfrac{x}2+5x+c$
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$\dfrac {x^4}4-\dfrac{x^4}3+5$

Explanation

Given

$\displaystyle \int x^2-x+5 \, dx$

$\implies \displaystyle \int x^2 dx-\int x dx+\int 5 dx$

$\implies \dfrac{x^3}3-\dfrac{x^2}2+5x+c$

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

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

Explanation

$\int x ^ { 3 } e ^ { x ^ { 2 } } d x$

put

$x ^ { 2 } = t \Rightarrow 2 x d x = d t$

$\dfrac { 1 } { 2 } \int t e ^ { t } d t = \dfrac { 1 } { 2 } \left[ \left( t e ^ { t } \right) - \int e ^ { t } d t \right]$ {integration by parts}

$= \dfrac { 1 } { 2 } \left[ t e ^ { t } - e ^ { t } \right]+c$

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

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

$\displaystyle\int x^2e^{x^3}\cos \left(e^{x^3}\right)dx$ is equal to?

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

Explanation

Let

$e^{x^3} = m$

$3x^2e^{x^3} dx = dm$

$\displaystyle \cfrac{1}{3} \int \cos m dm$

$\cfrac{1}{3} \sin e^{x^3}+C$

The value of

$\int \sqrt { \dfrac { e^ x\quad -\quad 1 }{ e^ x\quad +\quad 1 } } dx$ is equal to

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$\ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) - sec^ -1(e^ x)+c$
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$\ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) +sec^ -1(e^ x)+c$
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$\ell n\left( e^ x-\sqrt { e^ 2x\quad -1 } \right) -sec^ -1(e^ x)+c$
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$\ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) -sin^ -1(e^ {-x})+c$

Explanation

Solution

$\displaystyle\int \sqrt {\dfrac{e^{x}-1}{e^{x}+1}}dx$

$\Rightarrow \displaystyle\int \sqrt{\dfrac{e^{x}-1}{e^{x}+1}.\dfrac{e^{x}-1}{e^{x}+1}}dx$

$\Rightarrow \displaystyle \int \dfrac{e^{x}-1}{e^{2x}-1}dx$

$\Rightarrow \displaystyle \int\dfrac{e^{x}}{\left(e^{x}\right)^{2}-1}dx+\int \dfrac{-1}{\sqrt{e^{2x}-1}}dx$

$\Rightarrow \displaystyle \int\dfrac{e^{x}}{\left(e^{x}\right)^{2}-1}dx+\int \dfrac{-1e^{-x}}{1-\left(e^{-x}\right)^{2}}dx$

$=\left|e^{x}+\sqrt{e^{2x}-1}\right|+\sin^{-1}e^{-n}+c$

$\int \sqrt {1 + \sin x}dx =$

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$\dfrac {1}{2}\left (\sin \dfrac {x}{2} + \cos \dfrac {x}{2}\right ) + c$
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$\dfrac {1}{2}\left (\sin \dfrac {x}{2} - \cos \dfrac {x}{2}\right ) + c$
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$2\sqrt {1 + \sin x} + c$
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$-2\sqrt {1 - \sin x} + c$

Explanation

$\displaystyle\int \sqrt{1+\sin x} dx$

$=\displaystyle\int \dfrac{\sqrt{1+\sin x}.\sqrt{1-\sin x}}{\sqrt{1-\sin x}}\ dx$

$=\displaystyle\int \dfrac{\sqrt{(1+\sin x)(1-\sin x)}}{\sqrt{1-\sin x}}dx$

$=\displaystyle\int\dfrac{\sqrt{1-\sin^{2}x}}{\sqrt{1-\sin x}}dx=\int \dfrac{\cos x}{\sqrt{1-\sin x}}dx$

Let

$u=\sqrt{1-\sin x}$

$du=\dfrac{1}{\sqrt{1-\sin x}}(-\cos x)dx$

$-2du=\dfrac{\cos x}{2\sqrt{1-\sin x}}dx$

$=\displaystyle\int -2du=-2u+c$

$=-2\sqrt{1-\sin x}+c$

The integral of

$\displaystyle\int e^{\sin x}(x\cos x-\sec x\tan x)dx$ is?

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$se^{\sin x}-e^{\sin x}\sec x+c$
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$(x+\sec x)e^{\sin x}+c$
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$e^{\sin x}\cos x+c$
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$e^{\sin x}(\cos x-\sec x)+c$

The value of

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

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

Explanation

$\displaystyle I = \int (x-1)e^{-x}$

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

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

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

$\displaystyle = -xe^{-x}+c$

1097416_1134804_ans_85cdafbd5a5a477aa5ad92decf14c398.jpg

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

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$\dfrac {1}{2}\left [\tan^{-1}(x + 1) + \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
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$\dfrac {1}{2}\left [\tan^{-1}(x + 2) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
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$\dfrac {1}{2}\left [\tan^{-1}(x + 1) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
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$\dfrac {1}{2}\left [\tan^{-1}(x + 2) + \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$

$\int\dfrac{e^x+e^{-x}+(e^x-e^{-x})sin x}{1+cos x}dx=$

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$(e^x+e^{-x}) tan (x/2)+C$
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$(e^x-e^{-x}) cot (x/2)+C$
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$(e^x-e^{-x}) tan (x/2)+C$
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$(e^x-e^{-x}) cosec (x/2)+C$

$\displaystyle \int \dfrac{x^4 + 1}{x^6 + 1} dx = \tan^{-1} (f(x)) -\dfrac{2}{3} \tan^{-1} (g(x)) + C$ , then

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Both $f(x)$ & $g(x)$ are odd functions
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$g(x)$ is monotonic function
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none of these
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None

Explanation

$\int \dfrac{x^4+1}{x^6+1}dx$

$\int \dfrac{x^4+1}{x^6+1}\times \dfrac{x^2+1}{x^2+1} dx$

$\int \dfrac{(x^6+1)+x^2(x^2+1)}{(x^6+1)(x^2+1)}$

$\int \left ( \dfrac{1}{x^2+1} +\dfrac{1}{3}\dfrac{3x^2}{x^6+1}\right )dx$

$\tan^{-1}x+\dfrac{1}{3}\tan^{-1}(x^3)+C$

Integral of

$f ( x ) = \sqrt { 1 + x ^ { 2 } }$ with respect to

$x ^ { 2 }$ is

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$\frac { 2 } { 3 } \frac { \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } } { x } + k$
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$\frac { 2 } { 3 } \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k$
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$\frac { 2 } { 3 } x \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k$
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None of these

Explanation

$f(x)=\sqrt{1+x^2}$

$\int \sqrt{1+x^2} dx^2$

$=\dfrac{(1+x^2)^{\frac{3}{2}}}{\frac{3}{2}}$

$=\dfrac{2}{3}(1+x^2)^{\frac{3}{2}}+k$

The value of

$\displaystyle \int {\dfrac{{dx}}{{\sin x.\sin \left( {x + \alpha } \right)}}}$ equal to

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$\csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sin x}}{{\sin \left( {x + \alpha } \right)}}} \right| + c$
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$\;\;\;\;\csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sin \left( {x + \alpha } \right)}}{{\sin x}}} \right| + c$
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$\csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sec \left( {x + \alpha } \right)}}{{\sec x}}} \right| + c\;$
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$\,\;\;\csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sec x}}{{\sec \left( {x + \alpha } \right)}}} \right| + c$

Explanation

Given

$I= \displaystyle\int \dfrac{1}{sin x. sin(x+\alpha )}dx$

$=\dfrac{1}{sin\alpha } \displaystyle\int \dfrac{sin(x+\alpha -x)}{sin x. sin(x+\alpha )}dx$

$=\dfrac{1}{sin\alpha } \displaystyle\int \dfrac{sin(x+\alpha ).cos x-cos(x+\alpha )sin x}{sin x. sin (x+\alpha )}dx$

$=\dfrac{1}{sin\alpha } \displaystyle\int (cot x- cot (x+\alpha ))dx$

$=\dfrac{1}{sin\alpha }[log|sinx|-log |sin (x+\alpha )|]+c$

$= cosec \, \alpha \, log \left|\dfrac{sin x}{sin (x+\alpha )}\right|+c$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 7 - 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 7 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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