MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 5

$\displaystyle \int x^{3}\tan ^{-1}x\:dx.$

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$\displaystyle \frac{1}{2}\left [ \left ( x^{4}-1 \right )\tan ^{-1}x-\frac{x^{2}}{3}+x \right ].$
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$\displaystyle \frac{1}{2}\left [ \left ( x^{4}+1 \right )\tan ^{-1}x-\frac{x^{2}}{3}+x \right ].$
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$\displaystyle \frac{1}{4}\left [ \left ( x^{4}-1 \right )\tan ^{-1}x-\frac{x^{2}}{3}+x \right ].$
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$\displaystyle \frac{1}{4}\left [ \left ( x^{4}-1 \right )\tan ^{-1}x-\frac{x^{1}}{3}+x \right ].$

Explanation

Integrating by parts,

$\displaystyle I=\frac{x^{4}}{4}\tan ^{-1}x-\frac{1}{4}\int \frac{x^{4}-1+1}{1+x^{2}}dx.$

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

$\displaystyle =\frac{1}{4}\left [ \left ( x^{4}-1 \right )\tan ^{-1}x-\frac{x^{2}}{3}+x \right ]+c.$

$\displaystyle\int { x\left( f\left( { x }^{ 2 } \right) g''\left( { x }^{ 2 } \right) -f''\left( { x }^{ 2 } \right) g\left( { x }^{ 2 } \right) \right) dx } =$

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$\displaystyle f\left( { x }^{ 2 } \right) g'\left( { x }^{ 2 } \right) -g\left( { x }^{ 2 } \right) f'\left( { x }^{ 2 } \right) +c$
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$\displaystyle \frac { 1 }{ 2 } \left( f\left( { x }^{ 2 } \right) g\left( { x }^{ 2 } \right) f'\left( { x }^{ 2 } \right) \right) +c$
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$\displaystyle \frac { 1 }{ 2 } \left( f\left( { x }^{ 2 } \right) g'\left( { x }^{ 2 } \right) -g\left( { x }^{ 2 } \right) f'\left( { x }^{ 2 } \right) \right) +c$
0%
none of these

Explanation

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

Substitute

$x^2=t\Rightarrow 2xdx=dt$

$\displaystyle I=\frac { 1 }{ 2 } \int { \left( f\left( t \right) g''\left( t \right) -g\left( t \right) f''\left( t \right) \right) dt }$

$\displaystyle =\frac { 1 }{ 2 } \int { f\left( t \right) g''\left( t \right) dt } -\frac { 1 }{ 2 } \int { g\left( t \right) f''\left( t \right) }$

$\displaystyle =\frac { 1 }{ 2 } \left( f\left( t \right) g'\left( t \right) -\int { f'\left( t \right) g'\left( t \right) dt } -g\left( t \right) f'\left( t \right) +\int { g'\left( t \right) f'\left( t \right) dt } \right) +c$

$\displaystyle =\frac { 1 }{ 2 } \left( f\left( t \right) g'\left( t \right) -g\left( t \right) f'\left( t \right) \right) +c$

$\displaystyle =\frac { 1 }{ 2 } \left( f\left( { x }^{ 2 } \right) g'\left( { x }^{ 2 } \right) -g\left( { x }^{ 2 } \right) f'\left( { x }^{ 2 } \right) \right) +c$

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

$\displaystyle f\left ( 0 \right )=\frac{1+\sqrt{2}}{2}$ , then

$f(1)$ is equal to

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$\displaystyle \log \left (\sqrt{ \sqrt{2}-1} \right )$
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$\displaystyle \frac {-1}{\sqrt {2}}$
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$\displaystyle 1+ \sqrt{2}$
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$\displaystyle \dfrac12\log \left ( 1+\sqrt{2} \right )$

Explanation

By rationalizing the integrand, the given integral can be written as

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

$f(x)\displaystyle =-\frac { { x }^{ 2 } }{ 2 } -\frac { x }{ 2 } \sqrt { { x }^{ 2 }+1 } -\dfrac12\log { \left| x+\sqrt { { x }^{ 2 }+1 } \right| } +c$

Given,

$f(0) = \dfrac{1+\sqrt2}{2}+c$

$\therefore c = \dfrac{1+\sqrt2}{2}$

And,

$\displaystyle f(1) =-\frac { 1 }{ 2 } -\frac {\sqrt2}{ 2 } -\dfrac12\log { \left| 1+\sqrt { 2 } \right| } +\left(\frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 2 }}\right)$

$\displaystyle f(1)=-\dfrac12\log { \left| 1+\sqrt { 2 } \right| } =\dfrac12\log { \left( \sqrt { 2 } -1 \right) }$

$f(1) = \log{\left(\sqrt{\sqrt2 - 1}\right)}$

$\displaystyle I = \int \frac {\log (\cos x)}{\cos^2 x}dx$ , then

$I$ equals

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$\displaystyle \tan x \log \cos x + \tan x - x + C$
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$\displaystyle \tan x \log \cos x - \tan x - x^2 + C$
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$\displaystyle \tan x \log \cos x - \cot x + x + C$
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$\displaystyle \tan x \log \cos x + \cot x - x + C$

Explanation

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

$\displaystyle \int e^{\sin x}\left ( \frac{x\cos ^{3}x-\sin x}{\cos ^{2}x} \right )dx$

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$\displaystyle xe^{\sin x}-e^{\sin x}.\sec x$
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$\displaystyle e^{\sin x}-e^{\sin x}.\sec x$
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$\displaystyle xe^{\cos x}-e^{\sin x}.\sec x$
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$\displaystyle xe^{\sin x}+e^{\sin x}.\sec x$

Explanation

$\displaystyle I=\int \left [ x\left ( e^{\sin x}\cos x \right )-e^{\sin x}\left ( \sec x\tan x \right ) \right ]dx$
Integrate by parts.

$\displaystyle \left [ xe^{\sin x} -\int e^{\sin x}.1.dx\right ]-\left [ e^{\sin x}\sec x-\int e^{\sin x}\cos x\sec x\:dx \right ]$

$\displaystyle =xe^{\sin x}-e^{\sin x}.\sec x$ other integrals cancel

$\displaystyle I = \int \cos \theta \log (\tan \theta/2) d\theta$ , then I equals

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$\displaystyle \sin \theta \log (\tan \theta/2) + \theta + C$
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$\displaystyle \cos \theta \log (\tan \theta/2) + \theta + C$
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$\displaystyle \sin \theta \log (\tan \theta/2) - \theta + C$
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none of these

Explanation

$\displaystyle \int { \cos { \theta } \log { \left( \tan { \dfrac { \theta }{ 2 } } \right) d\theta } }$

$\displaystyle =\sin { \theta } \log { \left( \tan { \dfrac { \theta }{ 2 } } \right) } -\int { 2\sin { \dfrac { \theta }{ 2 } } \cos { \dfrac { \theta }{ 2 } } .\dfrac { \sec ^{ 2 }{ \dfrac { \theta }{ 2 } } }{ \tan { \dfrac { \theta }{ 2 } } } {\dfrac{1}{2}}d\theta } .$

$\displaystyle =\sin { \theta } \log { \left( \tan { \dfrac { \theta }{ 2 } } \right) } -\int { d\theta }$

$\displaystyle =\sin { \theta } \log { \left( \tan { \dfrac { \theta }{ 2 } } \right) } -\theta +C$

$\displaystyle \int \log \left ( x^{2}+x \right )dx=x\log x+\left ( x+1 \right )\log \left ( x+1 \right )+K,$ then K is equal to

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$\displaystyle 2x+\log\left ( x+1 \right )+C$
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$\displaystyle 2x-\log\left ( x+1 \right )+C$
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constant
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none of these

Explanation

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

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

$=x\log { \left( { x }^{ 2 }+x \right) } -2x+\log { \left( x+1 \right) } +c\\ =x\log { x } +\left( x+1 \right) \log { \left( x+1 \right) } -2x+c\\ \therefore k=-2x+c$

$\displaystyle \int e^{\cos x}\frac{\left ( x\sin^{3}x+\cos x \right )}{\sin ^{2}x}dx$

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$\displaystyle x.e^{\cos x}-e^{\cos x}co\sec x.$
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$\displaystyle -.e^{\cos x}+e^{\cos x}co\sec x.$
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$\displaystyle -x.e^{\cos x}-e^{\cos x}co\sec x.$
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$\displaystyle -x.e^{\cos x}+e^{\cos x}co\sec x.$

Explanation

Let

$\displaystyle I=\int { \frac { { e }^{ \cos { x } }\left( x\sin ^{ 3 }{ x } +\cos { x } \right) }{ \sin ^{ 2 }{ x } } } dx$

$=\int { { e }^{ \cos { x } } } \left( x\sin { x } +\cot { x } \csc { x } \right) dx\\ =\int { x{ e }^{ \cos { x } } } \sin { x } dx+\int { { e }^{ \cos { x } } } \csc { x } \cot { x } dx\\ ={ I }_{ 1 }+{ I }_{ 2 }$
Where

${ I }_{ 1 }=\int { x } { e }^{ \cos { x } }\sin { x } dx\\ =x\int { { e }^{ \cos { x } } } \sin { x } dx-\int { 1 } \left( \int { { e }^{ \cos { x } }\sin { x } dx } \right) dx\\ =-x{ e }^{ \cos { x } }+\int { { e }^{ \cos { x } } } dx+c$
And

${ I }_{ 2 }=\int { { e }^{ \cos { x } } } \csc { x } \cot { x } dx\\ ={ e }^{ \cos { x } }\int { \csc { x } } \cot { x } dx\\ -\int { \left( { e }^{ \cos { x } }\left( -\sin { x } \right) \right) } \left( \int { \csc { x } \cot { x } dx } \right) dx\\ ={ e }^{ \cos { x } }\left( -\csc { x } \right) +\int { { e }^{ \cos { x } }\sin { x } \left( -\csc { x } \right) } dx\\ =-{ e }^{ \cos { x } }\csc { x } -\int { { e }^{ \cos { x } }dx } \\ \therefore { I }_{ 1 }+{ I }_{ 2 }=-{ xe }^{ \cos { x } }+\int { { e }^{ \cos { x } }dx } -{ e }^{ \cos { x } }dx+c\\ =-{ e }^{ \cos { x } }\left( x+\csc { x } \right) +c$

$\displaystyle \int { { 2 }^{ x }\left( f'(x)+f(x)\log { 2 } \right) dx }$ is

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${ 2 }^{ x }f'(x)+C$
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${ 2 }^{ x }f(x)+C$
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${ 2 }^{ x }\left( \log { 2 } \right) f(x)+C$
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$\left( \log { 2 } \right) f(x)+C$

Explanation

Let

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

$=\displaystyle \int { { 2 }^{ x }f'(x)dx } +\int { { 2 }^{ x }f(x)dx } \log { 2 } dx$

$={ 2 }^{ x }f(x)+C$

If the graph of the antiderivative F(x) of f(x) = log (logx) + $(logx)^{-2}$ passes through (e. 1998\, -\, e) then the term independent of x in F(x) is .......... .

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1998
0%
1999
0%
1997
0%
1996

Explanation

An antiderivative of

$f(x) = F(x)$

$=\, \int (log(logx)\, +\, (log x)^{-2} dx\, +\, C$ (intergrating by parts the first term)

$=\, xlog(logx)\, -\, [x(logx)^{-1}\, +\, \int(log x)^{-2}\, dx]\, +\, \int(log\, x)^{-2}\, dx\, +\, C$ (again integrating by parts)

$=\, x log(log x)\, -\, x(logx)^{-1}\, +\, C$
Putting

$x = e$ , we have

$1998 - e = e.0 + e + C.$ Thus

$C = 1998$ .

$\displaystyle \int { \frac { { x }^{ 2 } }{ { \left( x\sin { x } +\cos { x } \right) }^{ 2 } } dx } =\frac { f\left( x \right) }{ x\sin { x } +\cos { x } } +\tan { x } +c$ , then

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$\displaystyle f\left( x \right)=\frac { x }{ \cos { x } }$
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$\displaystyle f\left( x \right) =\frac { \cos { x } }{ x }$
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$\displaystyle f\left( x \right)=-\frac { x }{ \cos { x } }$
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none of these

Explanation

Substitute

$\displaystyle \frac { 1 }{ x\sin { x } +\cos { x } } =z\Rightarrow -\frac { 1 }{ { \left( x\sin { x } +\cos { x } \right) }^{ 2 } } dx=dz$

$\displaystyle \Rightarrow \frac { x\cos { x } }{ { \left( x\sin { x } +\cos { x } \right) }^{ 2 } } =-dz\Rightarrow \int { \frac { { x }^{ 2 } }{ { \left( x\sin { x } +\cos { x } \right) }^{ 2 } } dx } =-z$

Now,

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

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

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

$\displaystyle =\frac { -x }{ \cos { x } \left( x\sin { x } +\cos { x } \right) } +\tan { x } +c$

$\displaystyle \therefore f\left( x \right) =-\frac { x }{ \cos { x } }$

$\int\displaystyle\frac{x^2+1}{x^4+1}dx$ is equal to.

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$\displaystyle\frac{1}{\sqrt 2}\tan^{-1}\left(\displaystyle\frac{x^2}{\sqrt 2 x}\right)+c$
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$\displaystyle\frac{1}{\sqrt 2}\tan^{-1}\left(\displaystyle\frac{x^2-1}{\sqrt 2 x}\right)+c$
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$\displaystyle\frac{1}{\sqrt 2}\tan^{-1}\left(\displaystyle\frac{x^2+1}{x}\right)+c$
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$\displaystyle\tan^{-1}\left(\displaystyle\frac{x^2-1}{\sqrt 2 x}\right)+c$

Explanation

Let

$I=\displaystyle\int\frac{x^2+1}{x^4+1}dx$

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

$=\displaystyle\int\frac{dt}{t^2+(\sqrt 2)^2}$ , where

$t=x-\frac{1}{x}$

$=\displaystyle\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{t}{\sqrt 2}\right)+c=\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x^2-1}{\sqrt 2x}\right)+c$

Evaluate

$\displaystyle \int xe^{x} dx$ .

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

Explanation

Let

$I = \displaystyle \int xe^{x} dx$

Using integration by parts,

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

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

$\displaystyle I = xe^x-\int 1 \cdot e^xdx$

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

Identify the correct expression :

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$x \int lnx\, dx = x^{2} ln |x| - x^{2} + c$
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$x \int ln |x| dx = xe^{x} + c$
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$x \int e^{x} dx = xe^{x} + c x$
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$\displaystyle \int \frac{dx}{\sqrt{a^{2} + x^{2}}} = \frac{1}{a} tan^{-1} \left ( \frac{x}{a} \right ) + c$

Explanation

$\displaystyle x\int { \ln { x } dx } =x\left( x\ln { x } -\int { \frac { x }{ x } dx } \right)$

$=x\left( x\ln { x } -x+c \right) ={ x }^{ 2 }\ln { x } -{ x }^{ 2 }+cx\\ x\int { { e }^{ x }dx } =x\left( { e }^{ x }+c \right) =x{ e }^{ x }+cx$

$\displaystyle \int { \frac { dx }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } } =\log { \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) } +c$

$\displaystyle \int{\tan^2x}dx$

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$\tan x-x+c$
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$\tan x+c$
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$\tan x-x$
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None of the above

Explanation

$\displaystyle \int{\tan^2x}dx=\int(\sec^2x-1)dx=\int \sec^2xdx-\int dx = \tan x-x+C$

Primitive of

$\displaystyle \frac{3x^4\, -\, 1}{(x^4\, +\, x\, +\, 1)^2}$ w.r.t. x is

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$\displaystyle \frac{x}{x^4\, +\, x\, +\, 1}\, +\, c$
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$\displaystyle - \frac{x}{x^4\, +\, x\, +\, 1}\, +\, c$
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$\displaystyle \frac{x\, +\, 1}{x^4\, +\, x\, +\, 1}\, +\, c$
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$\displaystyle -\frac{x\, +\, 1}{x^4\, +\, x\, +\, 1}\, +\, c$

Explanation

$\displaystyle \int \frac{3x^4\, -\, 1}{(x^4\, +\, x\, +\, 1)^2}dx\, =\, -\, \int \left ( \frac{1}{x^4\, +\, x\, +\, 1}\, -\, \frac{x(4x^3\, +\, 1)}{(x^4\, +\, x\, +\, 1)^2}\right ) dx$

$=\displaystyle \frac{-x}{x^4\, +\, x\, +\, 1}\, +\, c$ ,

Using $\displaystyle \left ( \int [f(x)\, +\, xf`\, (x)]dx\, =\, xf(x)\, +\, C\right )$

Evaluate

$\displaystyle \int (\tan(e^{x}) + xe^{x}. \sec^{2}(e^{x}))dx$

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$x\, -\tan(e^{x}) + c$
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$x\, \tan(e^{x}) + c$
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$x\, \tan(e^{-x}) + c$
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$x\, -\tan(e^{-x}) + c$

Explanation

$I = \displaystyle \int (\tan(e^{x}) + xe^{x}. \sec^{2}(e^{x}))dx$

$= \int \tan(e^x)dx+\int xe^{x}. \sec^{2}(e^{x})dx=I_1+I_2$

$I_1 =\displaystyle\int 1.\tan(e^x)dx$

We have,

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

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

$\displaystyle I_1= x \tan(e^x)-\int \frac{d\{ \tan(e^x)\}}{dx} x.dx+c$ , Using integral by parts.

$\displaystyle \quad = x \tan(e^x)-\int xe^{x}. \sec^{2}(e^{x})dx +c = x \tan(e^x)-I_2+c$

$\Rightarrow I = I_1+I_2 = x \tan(e^x) +c$

$\dfrac {dy}{dt} = ky$ and

$k\neq 0$ , which of the following could be the equation of

$y$ ?

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$y = kx - 7$
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$y = 95e^{kt}$
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$y = 5 + ln\ k$
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$y = (x - k)^{2}$
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$y = \sqrt [k]{x}$

Explanation

Given,

$\dfrac{dy}{dt} = ky$ and

$k \ne 0$
We have

$\dfrac{dy}{y} = kdt$
Apply integral on both sides, so we get

$\displaystyle \int \dfrac{dy}{y} = \int kdt$

$\Rightarrow \ln(y) = kt+c$ (where

$c$ is constant)

$\Rightarrow y = {e}^{c}{e}^{kt} = a{e}^{kt}$ (

$a = {e}^{c}$ )
The possible solution is

$y = 95{e}^{kt}$

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

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$(x + 1)e^{x + x^{-1}} + C$
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$(x - 1)e^{x + x^{-1}} + C$
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$xe^{x + x^{-1}} + C$
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$xe^{x + x^{-1}} x + C$

Explanation

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

$\displaystyle = \int [x.e^{x + x^{-1}} \left (1 - \dfrac {1}{x^{2}}\right ) + e^{x + x^{-1}}]dx$

$\left [\displaystyle \because \int xf'(x) + f(x) dx = xf(x) + C \right]$

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

$\displaystyle \int { \cfrac { f(x) }{ \log { (\sin { x } ) } } } dx=\log { \left[ \log { \sin { x } } \right] } +c$ , then

$f(x)=$ ........

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$\cot { x }$
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$\tan { x }$
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$\sec { x }$
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$co\sec { x }$

Explanation

Here,

$\displaystyle \int { \dfrac { f(x) }{ \log (\sin x) } dx } = \log (\log (\sin x))+c$

R.H.S.=

$\log(\log (\sin x))$

Differentiating the R.H.S

$\Rightarrow$

$\dfrac { 1 }{ \log (\sin x) } \times \dfrac { 1 }{ \sin x } \times \cos x=\dfrac { \cot x }{ \log (\sin x) }$

$\Rightarrow f(x)=\cot x$

$\int \dfrac {1}{x^{2} (x^{4} + 1)^{3/4}}dx$ is equal to

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$\left (1 + \dfrac {1}{x^{4}}\right )^{1/4} + C$
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$(x^{4} + 1)^{1/4} + C$
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$\left (1 - \dfrac {1}{x^{4}}\right )^{1/4} + C$
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$-\left (1 + \dfrac {1}{x^{4}}\right )^{1/4} + C$

Explanation

$\int \dfrac {dx}{x^{2}(x^{4} + 1)^{3/4}} = \int \dfrac{dx}{x^2 \cdot x^3\left(\dfrac{x^4+1}{x^4}\right)^{3/4}}=\int \dfrac {dx}{x^{5}\left (1 + \dfrac {1}{x^{4}}\right )^{3/4}}$
Put

$1 + \dfrac {1}{x^{4}} = t\Rightarrow -\dfrac {4}{x^{5}}dx = dt$
So, integral is

$I = -\dfrac {1}{4}\int \dfrac {dt}{t^{3/4}} =-\dfrac{1}{4}\dfrac{(t)^{\frac{1}{4}}}{\frac{1}{4}}+c= -t^{\frac {1}{4}} + c = -\left (1 + \dfrac {1}{x^{4}}\right )^{1/4} + c$

Hence, option D is correct.

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

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$\log (1+cos^2x)+C$
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$\log (1+tan^2x)+C$
0%
$-\log (1+sin^2x)+C$
0%
$-\log (1+cos^2x)+C$

Explanation

Let

$I = \displaystyle \int \frac{\sin 2x}{\sin^2x+2 \cos^2x}dx=\int \dfrac{\sin 2x}{(\sin^2x+\cos^2x)+\cos^2x}dx$

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

Substitute

$1+\cos^2x=u$

$\Rightarrow 0+2\cos x(-\sin x)dx=du$

$\Rightarrow \sin 2xdx=-du$

Therefore

$\displaystyle I=-\int \dfrac{du}{u}=-\ln u+c=-\ln(1+\cos^2x)+c$

What is

$\displaystyle \int \dfrac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} dx$ equal to?

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$\sqrt{\dfrac{x^4 + x^2 + 1}{x}} + c$
0%
$\sqrt{x^4 + 2 - \dfrac{1}{x^2}} + c$
0%
$\sqrt{x^2 + \dfrac{1}{x^2} + 1} + c$
0%
$\sqrt{\dfrac{x^4 - x^2 + 1}{x}} + c$

Explanation

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

$=\displaystyle \int \dfrac{x^4 - 1}{x^3 \sqrt{x^2 + 1 + \dfrac{1}{x^2}}} dx$

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

Put

$x^2 + 1 + \dfrac{1}{x^2}=t$

$\therefore \left(2x-\dfrac{2}{x^3}\right)dx=dt$

$\therefore \left(x-\dfrac{1}{x^3}\right)dx=\dfrac{dt}{2}$

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

$=\sqrt t +C$

$=\sqrt{x^2 + 1 + \dfrac{1}{x^2}}+C$

Answer is option (C)

What is

$\displaystyle\int { \left( x\cos { x } +\sin { x } \right) dx }$ equal to?

Where

$c$ is an arbitrary constant

0%
$x\sin { x } +c$
0%
$x\cos { x } +c$
0%
$-x\sin { x } +c$
0%
$-x\cos { x } +c$

Explanation

$\int { (x.\cos x+\sin x) } dx$

$=\int { x.\cos x.dx } +\int { \sin x.dx }$

By u\sin g uv rule for integration in the first integral taking x as first function and \cos x as second we have,

$=x.\sin x-\int { \sin x.dx } +\int { \sin x.dx }+c$

$=x.\sin x+c$

Hence, A is correct.

$\int { \left( \cfrac { 4{ e }^{ x }-25 }{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$ , then

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$A=5,B=3$
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$A=5,B=-3$
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$A=-5,B=3$
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$A=-5,B=-3$

Explanation

Given :

$\displaystyle \int { \left( \cfrac { 4{ e }^{ x }-25 }{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$

$\implies \displaystyle \int { \left( \cfrac { 10 e^{x} -6{ e }^{ x }-25 }{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$

$\implies \displaystyle \int {\left(5 - \cfrac { 6{e}^{x}}{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$

$\implies \displaystyle \int 5. dx - 3\int {\left(\cfrac { 2{e}^{x}}{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$

$\implies 5x -3 \log|2e^{x}-5|+c=Ax+B\log|2e^{x}-5|+c$

Equating the like terms we get

$A=5$ and

$B=-3$

Hence, option B is correct.

The value of

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

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$\dfrac {e^x} {x} + C$
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$x e^x \log |x| + C$
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$e^x \log |x| + C$
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$x ( e^x + \log |x| ) + C$
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$xe^x + \log |x| + C$

Explanation

Let

$I=\displaystyle \int \dfrac {e^x}{x} ( x \log x+1) dx$
Therefore,

$I =\displaystyle \int \underset {II}{e^x} \underset {I}{\log} x dx + \int \dfrac {e^x}{x} dx$ ..................(taking

$e^x$ as 2nd function and

$log x$ as 1st function)

$= e^x \log x - \displaystyle \int \dfrac {e^x}{x} dx + \int \dfrac {e^x}{x} dx + C$ .........(applying by-parts integration formula)

$= e^x \log x + C$

As the log function is not defined for -ve

$x$

$\therefore \ the \ answer \ is \ e^x log |x|+c$

So the answer is option

$C$ .

$\displaystyle\int \displaystyle\frac{\cos \alpha}{\sin x\cos (x-\alpha)}dx=$ ___________

$+$ c where

$0 < x < \alpha < \pi_{/2}$ and

$\alpha$ -constant.

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$-ln|tan x+\cot \alpha|$
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$ln|\cot x+\tan\alpha|$
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$ln|tan x+\cot \alpha|$
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$-ln|\cot x+\tan\alpha|$

Explanation

$\int \dfrac{cos\alpha}{sinxcos(x-\alpha)}dx$

$=\int \dfrac{cos(x-(x-\alpha))}{sinxcos(x-\alpha)}dx$

$=\int \dfrac{cosxcos(x-\alpha)+sinxsin(x-\alpha)}{sinxcos(x-\alpha)}dx$

$=\int (cotx+tan(x-\alpha))dx$

$= ln|sinx|-ln|cos(x-\alpha)|$

$= ln|\dfrac{sinx}{cos(x-\alpha)}|$

$= ln|\dfrac{sinx}{cosxcos(\alpha)+sinxsin(\alpha)}|$

$=ln|\dfrac{sec(\alpha)}{cotx+tan(\alpha)}|$

$=ln|(sec(\alpha))-ln(cotx+tan(\alpha))|$

$=-ln|(cotx+tan(\alpha))|+C$

Hence, the answer is option (D).

$\int \dfrac {1}{x \sqrt {1 - x^{3}}}dx = a\log \left |\dfrac {\sqrt {1 - x^{3}} - 1}{\sqrt {1 - x^{3}} + 1}\right | + b$ then

$a =$

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$\dfrac {2}{3}$
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$\dfrac {1}{3}$
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$-\dfrac {2}{3}$
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None of these

Explanation

$I=\int { \cfrac { 1 }{ x\sqrt { 1-{ x }^{ 3 } } } dx } =\int { \cfrac { { x }^{ 2 } }{ { x }^{ 3 }\sqrt { 1-{ x }^{ 3 } } } dx }$

Let

$\sqrt { 1-{ x }^{ 3 } } =t$

$I=\int { \cfrac { 1 }{ x\sqrt { 1-{ x }^{ 3 } } } dx } =\int { \cfrac { { x }^{ 2 } }{ { x }^{ 3 }\sqrt { 1-{ x }^{ 3 } } } dx } \\ \Rightarrow \cfrac { 1 }{ 2\sqrt { 1-{ x }^{ 3 } } } x(-3{ x }^{ 2 })dx=dt\\ \Rightarrow I=\cfrac { 2 }{ 3 } \int { \cfrac { 1 }{ { t }^{ 2 }-1 } dt } \\ \Rightarrow \cfrac { 3 }{ 2 } I=\int { \cfrac { 1 }{ (t+1)(t-1) } dt } =\cfrac { 1 }{ 2 } \int { \cfrac { (t+1)-(t-1) }{ (t+1)(t-1) } } dt\\ \Rightarrow 3I=\int { \cfrac { 1 }{ (t-1) } dt } -\int { \cfrac { 1 }{ (t+1) } dt } \\ =\ln { \left| t-1 \right| } -\ln { \left| t+1 \right| } +b$

where

$b$ is a constant

$\Rightarrow 3I= \ln{ \left| \cfrac{t-1}{t+1} \right| }+b$

Converting

$t=\sqrt{1-x^3}$ , we get

$I=\cfrac{1}{3} \ln{ \left| \cfrac{\sqrt{1-x^3} -1}{\sqrt{1-x^3} +1} \right| }$

Hence, the correct answer is

$a=\cfrac{1}{3}$

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

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$-e^{-x} \text{cosec} x + C$
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$e^{-x} \text{cosec} x + C$
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$-e^{-x} (\text{cosec} x + \cot x) + C$
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$-e^{-x} (\text{cosec} x + \tan x) + C$
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$-e^{-x}\sec x + C$

Explanation

Let

$l = \int e^{-x} \text{cosec} x(1 + \cot x) dx$

$= \int e^{-x} \text{cosec} dx + \int e^{-x} \text{cosec} x\cdot \cot x dx$

$= \text{cosec} x \cdot (-e^{-x}) - \int (-\text{cosec} x\cdot \cot x) (-e^{-x})dx + \int e^{-x} \text{cosec} x \cdot \cot x dx$

$= -e^{-x} \cdot \text{cosec} x - \int e^{-x} \cdot \text{cosec} x \cdot \cot dx + \int e^{-x} \text{cosec} x \cdot \cot x dx$

$= -e^{-x} \text{cosec} x + c$

The solution of the equation

$\dfrac { dy }{ dx } =\dfrac { x\left( 2\log { x } +1 \right) }{ \sin { y } +y\cos { y } }$ is

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$y\sin { y } ={ x }^{ 2 }\log { x } +\dfrac { { x }^{ 2 } }{ y } +c$
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$y\cos { y } ={ x }^{ 2 }\left( \log { x } +1 \right) +c$
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$y\cos { y } ={ x }^{ 2 }\log { x } +\dfrac { { x }^{ 2 } }{ 2 } +c$
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$y\sin { y } ={ x }^{ 2 }\log { x } +c$

Explanation

Given equation is

$\dfrac {dy}{dx}=\dfrac {x(2\log x+1)}{\sin y+y \cos y}$

Therefore,

$\left( y\cos { y } +\sin { y } \right) dy=\left( 2x\log { x } +x \right) dx$

$\Rightarrow y\sin { y } -\displaystyle\int { \sin { y } dy } +\displaystyle\int { \sin { y } dx } ={ x }^{ 2 }\log { x } -\displaystyle\int { { x }^{ 2 }\cdot \dfrac { 1 }{ x } dx } +\displaystyle\int { xdx } +c$

$\Rightarrow y\sin { y } ={ x }^{ 2 }\log { x } +c$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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