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

If $$\displaystyle I=\int e^{-x}\log (e^{x}+1)dx$$, then I equals
  • $$\displaystyle x+(e^{-x}+1)\log (e^{x}+1)+C$$
  • $$\displaystyle x+(e^{x}+1)\log (e^{x}+1)+C$$
  • $$\displaystyle x-(e^{-x}+1)\log (e^{x}+1)+C$$
  • none of these
If $$I$$ $$=\displaystyle \int \sqrt {\frac {5-x}{2+x}}dx$$, then $$I$$ equals
  • $$\sqrt {x+2}\sqrt {5-x}+3 \sin^{-1}\sqrt {\dfrac {x+2}{3}}+C$$
  • $$\sqrt {x+2}\sqrt {5-x}+3 \sin^{-1}\sqrt {\dfrac {x+2}{7}}+C$$
  • $$\sqrt {x+2}\sqrt {5-x}+5 \sin^{-1}\sqrt {\dfrac {x+2}{5}}+C$$
  • None of these
$$\int \cos(\ln x)dx=$$
  • $$\dfrac{x}{2}(\cos \ln x+\sin \ln x)$$
  • $$\dfrac{x}{2}[\cos \ln x-\sin \ln x]$$
  • $$[x \cos \ln x + \sin \ln x]$$
  • None of these
The evaluation of $$\displaystyle \int \frac{pX^{p+2q-1}-qX^{q-1}}{X^{2p+2q}+2X^{p+q}+1}dx$$ is 
  • $$-\displaystyle \frac{X^p}{X^{p+q}+1}+C$$
  • $$\displaystyle \frac{X^q}{X^{p+q}+1}+C$$
  • $$-\displaystyle \frac{X^q}{X^{p+q}+1}+C$$
  • $$\displaystyle \frac{X^p}{X^{p+q}+1}+C$$
$$\displaystyle \int x\sin x\sec ^{3}x  dx$$ is equal to
  • $$\displaystyle \frac{1}{2}\left [ \sec^{2}x-\tan x \right ]+c$$
  • $$\displaystyle \frac{1}{2}\left [ x\sec^{2}x-\tan x \right ]+c$$
  • $$\displaystyle \frac{1}{2}\left [ x\sec^{2}x+\tan x \right ]+c$$
  • $$\displaystyle \frac{1}{2}\left [\sec^{2}x+\tan x \right ]+c$$
$$\displaystyle \int \frac{x^{2}\left ( x\sec^{2}x+\tan x \right )}{\left ( x\tan x+1 \right )^{2}}dx$$
  • $$\displaystyle x^{2}\left [ -\frac{1}{x\tan x+1} \right ]+2\log \left( x\sin x+\cos x \right)+C $$
  • $$\displaystyle x^{2}\left [ \frac{1}{x\tan x+1} \right ] +2\log \left( \sin x+x\cos x \right) +C$$
  • $$\displaystyle x\left [ -\frac{1}{x\tan x+1} \right ]+2\log \left( \sin x+x\cos x \right)+C $$
  • $$\displaystyle x\left [ \frac{1}{x\tan x+1} \right ] +2\log \left( x\sin x+\cos x \right) +C$$
Integrate $$\displaystyle x\sqrt{{a^{2}-x^{2}} }$$ with respect to $$x$$.
  • $$-(a^{2}-x^{2})+C$$ 
  • $$\displaystyle\sqrt{a^2-x^2}+C$$
  • $$-\displaystyle \dfrac{\left(a^2-x^2\right)^{\frac{3}{2}}}{3}+C$$ 
  • 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
  • $$\displaystyle\left(\frac{e^{\displaystyle 2\ln{x}}-1}{2}\right)e^{\displaystyle x^2}+C$$
  • $$\displaystyle\frac{(x-1)xe^{\displaystyle x^2}}{2}+C$$
  • $$\displaystyle\frac{(x^2-1)}{2x}-e^{\displaystyle x^2}+C$$
  • none of these
$$\displaystyle \int \sin 2x.\log \cos x\: dx$$ is equal to
  • $$\displaystyle \cos ^{2}x\left ( \frac{1}{2}+\log \cos x \right )+k$$
  • $$\displaystyle \cos ^{2}x.\log \cos \: x+k$$
  • $$\displaystyle \cos ^{2}x\left ( \frac{1}{2}-\log \cos x \right )+k$$
  • none of these
If $$\displaystyle \int { f(x) } dx=F(x)$$, then $$\displaystyle \int { { x }^{ 3 } } f\left( { x }^{ 2 } \right) dx$$ equals to
  • $$ \displaystyle \frac {1}{2} \left[x^2(F(x))^2 - \int (F(x))^2 dx \right]$$
  • $$ \displaystyle \frac {1}{2} \left[x^2F(x^2) - \int F(x^2) d(x^2) \right]$$
  • $$ \displaystyle \frac {1}{2} \left[x^2F(x) - \frac{1}{2} \int (F(x))^2 dx \right]$$
  • $$ \displaystyle \frac {1}{2} \left[x^2F(x^2) + \int F(x^2) d(x^2) \right]$$
If $$\displaystyle \int x\log \left ( 1+x^{2} \right )dx=\phi \left ( x \right ).\log \left ( 1+x^{2} \right )+\Psi \left ( x \right )+c$$ then
  • $$\displaystyle \phi \left ( x \right )=\frac{1+x^{2}}{2}$$
  • $$\displaystyle \Psi \left ( x \right )=\frac{1+x^{2}}{2}$$
  • $$\displaystyle \Psi \left ( x \right )=-\frac{1+x^{2}}{2}$$
  • $$\displaystyle \phi \left ( x \right )=-\frac{1+x^{2}}{2}$$
$$\displaystyle \int { \left( \sqrt { { x }^{ 2 }+1 }  \right) \left[ \frac { \ln { \left( { x }^{ 2 }+1 \right)  } -2\ln { x }  }{ { x }^{ 4 } }  \right] dx } ;g\left( x \right)=\left( 1+\frac { 1 }{ { x }^{ 2 } }  \right) ;$$ equals
  • $$\displaystyle \frac { 1 }{ 3 } { \left( g\left( x \right) \right)  }^{ 1/2 }\ln { \left( g\left( x \right) \right)  } +\frac { 2 }{ 9 } \left( g\left( x \right) \right) +C$$
  • $$\displaystyle -\frac { 1 }{ 3 } { \left( g\left( x \right)  \right)  }^{ 3/2 }\ln { \left( g\left( x \right)  \right)  } +\frac { 2 }{ 9 } { \left( g\left( x \right)  \right)  }^{ 3/2 }+C$$
  • $$\displaystyle \frac { 1 }{ 3 } { \left( g\left( x \right) \right)  }^{ 3/2 }\ln { \left( g\left( x \right) \right)  } -\frac { 2 }{ 9 } \left( g\left( x \right) \right) +C$$
  • None of these
$$\displaystyle \int \frac{e^{\tan^{-1} x}}{(1+x^2)}\left [\left ( \sec^{-1} \sqrt{1+x ^2} \right )^2+ \cos^{-1} \left ( \frac{1-x^2}{1+x^2} \right )dx \right ]$$ at $$(x>0)$$
  • $$e^{\tan ^{-1}x}.\tan^{-1} x+C$$
  • $$\displaystyle \frac{e^{\tan ^{-1}x}.\left ( \tan^{-1} x \right )^2}{2}+C$$
  • $$e^{\tan ^{-1}x}.\left ( \sec^{-1} \left ( \sqrt{1+x^2} \right )\right )^2+C$$
  • $$e^{\tan ^{-1}x}.\left ( \mathrm{cosec} ^{-1} \left ( \sqrt{1+x^2} \right )\right )^2+C$$
$$\displaystyle \int \left [{ \log  \left( 1+\cos  x \right) -x\tan  \left( \frac{x}{2} \right)  }\right ] dx$$
  • $$\displaystyle x\log\left ( 1+\tan x \right )$$
  • $$\displaystyle x\log\left ( 1+\sin x \right )$$
  • $$\displaystyle \log\left ( 1+\sin x \right )$$
  • $$\displaystyle x\log\left ( 1+\cos x \right )$$
$$\displaystyle\int \frac{2x+\sin 2x}{1+\cos 2x}dx$$ 
  • $$x\cot x.$$
  • $$x\tan x.$$
  • $$x^{2}\tan x.$$
  • $$x\sec x.$$
If $$\displaystyle I=\int_{0}^{1}\frac{e^{t}}{1+t}$$ dt, then $$\displaystyle p=\int_{0}^{1}e^{t}\log\left ( 1+t \right )dt=$$
  • $$I$$
  • $$2I$$
  • $$\displaystyle e\log2-I$$
  • none
Eavaluate :  $$\displaystyle  \int \displaystyle x\tan^{-1}x$$  
  • $$\displaystyle \frac{1}{2}(x^{2}+1)\tan ^{-1}x-\frac{1}{2}x+C$$
  • $$\displaystyle \frac{1}{2}x^{2}\tan ^{-1}x-\frac{1}{2}x+ C$$
  • $$\displaystyle \frac{1}{2}x^{2}\tan ^{-1}x+\frac{1}{2}x+C$$
  • none of these
If $$\displaystyle \int \left ( x^{3}-2x^{2}+3x-1 \right )\cos 2x\:dx$$
$$\displaystyle =\frac{\sin 2x}{4}u\left ( x \right )+\frac{\cos 2x}{8}v\left ( x \right )+C$$ then
  • $$\displaystyle u\left ( x \right )=x^{3}-4x^{2}+3x$$
  • $$\displaystyle u\left ( x \right )=2x^{3}-4x^{2}+3x$$
  • $$\displaystyle u\left ( x \right )=3x^{3}-4x+3$$
  • $$\displaystyle v\left ( x \right )=6x^{2}-8x$$
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
  • $$\displaystyle \frac {2n - 3}{2n - 1}$$
  • $$\displaystyle \frac {2n - 3}{2n - 2} \frac {a}{b}$$
  • $$\displaystyle \frac {n - 2}{n - 1} \frac {a}{b}$$
  • $$\displaystyle \frac {n - 3}{n - 2} \frac {a}{b}$$
If $$\displaystyle I=\int \sec^{2}x\:cosec ^{4}\:x\:dx= K\cot^{3}x+L\tan x+M\cot x+C$$ then
  • K$$=-1/3$$
  • L$$=2$$
  • M$$=-2$$
  • none of these
If $$\displaystyle \int \frac {xe^x}{\sqrt {1 + e^x}} dx = f(x) \sqrt {1 + e^x} -2 \log \: g(x) + C$$, then
  • $$\displaystyle f(x) = x - 1$$
  • $$\displaystyle g(x) = \frac {\sqrt {1 + e^x} - 1}{\sqrt {1 + e^x} + 1}$$
  • $$\displaystyle g(x) = \frac {\sqrt {1 + e^x} + 1}{\sqrt {1 + e^x} - 1}$$
  • $$\displaystyle f(x) = 2(x - 2)$$
If $$\displaystyle \int \log (\sqrt {1 - x} + \sqrt {1 + x}) dx = xf(x) + Lx + M \: sin^{-1} x + C$$, then
  • $$\displaystyle f(x) = \log (\sqrt {1 - x} + \sqrt {1 + x})$$
  • $$\displaystyle L = - \frac {1}{2}$$
  • $$\displaystyle M = - \frac {2}{3}$$
  • $$\displaystyle M = - \frac {1}{2}$$
If $$\int (x^{3}\, -\, 2x^{2}\, +\, 3x\, -\, 1)\, cos2x\, dx\, =\, \displaystyle \frac{sin 2x}{4}u(x)\, +\, \frac{cos 2x}{8}v(x)\, +\, c$$, then
  • $$u(x)\, =\, x^{3}\, -\, 4x^{2}\, +\, 3x$$
  • $$u(x)\, =\, 2x^{3}\, -\, 4x^{2}\, +\, 3x$$
  • $$v(x)\, =\, 3x^{2}\, -\, 4x\, +\, 3$$
  • $$v(x)\, =\,k 6x^{2}\, -\, 8x$$
$$\int \left [\displaystyle \frac {\sqrt{x^2\, +\, 1}\, [ln(x^2\, +\, 1)\, -\, 2\, ln\, x]}{x^4} \right ]$$
  • $$\displaystyle \frac {(x^2\, -\, 1)\, \sqrt{x^2\, -\, 1}}{9x^3}\, \left [2\, -\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$$
  • $$\displaystyle \frac {(x^2\, -\, 1)\, \sqrt{x^2\, -\, 1}}{9x^3}\, \left [2\, +\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$$
  • $$\displaystyle \frac {(x^2\, +\, 1)\, \sqrt{x^2\, +\, 1}}{9x^3}\, \left [2\, -\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$$
  • $$\displaystyle \frac {(x^2\, -\, 1)\, \sqrt{x^2\, +\, 1}}{9x^3}\, \left [2\, +\, 3\, ln\, \left (1\, +\, \displaystyle \frac {1}{x^2} \right ) \right ]$$
If $$\displaystyle I = \int \frac {\sec^{-1} \sqrt x - cosec^{-1} \sqrt x}{\sec^{-1} \sqrt x + cosec^{-1} \sqrt x} dx$$, then I equals
  • $$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) + x + C$$
  • $$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x + \sqrt {x - 1}) - x + C$$
  • $$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) - x + C$$
  • none of these
If $$\displaystyle \int xe^{-5x^2} \: sin \: 4x^2 dx = Ke^{-5x^2} (A \: sin \: 4x^2 + B \: cos \: 4x^2) + C$$. Then
  • $$\dfrac {1}{82}, 5, 4$$
  • $$\dfrac {1}{82}, -5, 4$$
  • $$\dfrac {1}{82}, -5, -4$$
  • $$\dfrac {1}{82}, \dfrac{-1}{5}, \dfrac{-1}{4}$$
$$\int e^{x^4}\, (x\, +\, x^3\, +\, 2x^5)e^{x^2}dx$$ is equal to
  • $$\displaystyle \frac{1}{2} xe^{x^2}\, \cdot\, e^{x^4}\, +\, c$$
  • $$\displaystyle \frac{1}{2} x^2\, e^{x^4}\, +\, c$$
  • $$\displaystyle \frac{1}{2} e^{x^2}\, \cdot\, e^{x^4}\, +\, c$$
  • $$\displaystyle \frac{1}{2} x^2 e^{x^2}\, \cdot\, e^{x^4}\, +\, c$$
If $$\displaystyle  I = \int \left ( \frac {1}{x^{1/2} - x^{1/4}} + \frac {\log (1 + x^{1/4})}{x^{1/2} + x^{1/4}} \right ) dx$$, then I equals
  • $$\displaystyle 2 \sqrt x + 4x^{1/4} + \log |x^{1/4} - 1| + C$$
  • $$\displaystyle 2(x^{1/2} - 6x^{1/4} + 3) \log (x^{1/4} + 1) + C - (x^{1/2} - 10x^{1/4} + 7) + 2(\log (x^{1/4} + 1))^2$$
  • $$\displaystyle \sqrt x - x^{1/4} + 6 \log (x^{1/4} + 1) - 3x^{3/4} + C$$
  • none of these
$$\displaystyle \int\, sin^2\, (lnx)\, dx$$ is equal to
  • $$\displaystyle \frac{2x}{15} (5 + 2sin(2lnx) + cos(2lnx)) + c$$
  • $$\displaystyle \frac{x}{5} (5 + 2sin(2lnx) -cos(2lnx)) + c$$
  • $$\displaystyle \frac{x}{10} (5 -2sin(2lnx)-  cos(2lnx)) + c$$
  • $$\displaystyle \frac{x}{10} (5 -2sin(2lnx) + sin(2lnx)) + c$$ 
$$\displaystyle\int\, \cos\, 2\theta.\,\ln\, \displaystyle \frac {\cos\theta\, +\, \sin\, \theta}{\cos\, \theta\, -\, \sin\, \theta}\, d\theta$$
  • $$\, \displaystyle \frac {1}{2}\, ln\, \left ( \displaystyle \frac {\cos\, \theta\, -\, \sin\, \theta}{\cos\, \theta\, +\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\sec\, 2\,\theta)\, +\, C$$
  • $$\, \displaystyle \frac {1}{2}\, \ln\, \left ( \displaystyle \frac {\cos\, \theta\, -\, \sin\, \theta}{\cos\, \theta\, +\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\text{cosec}\, 2\,\theta)\, +\, C$$
  • $$\, \displaystyle \frac {1}{2}\, \ln\, \left ( \displaystyle \frac {\cos\, \theta\, +\, \sin\, \theta}{\cos\, \theta\, -\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\text{cosec}\, 2\,\theta)\, +\, C$$
  • $$\, \displaystyle \frac {1}{2}\, \ln\, \left ( \displaystyle \frac {\cos\, \theta\, +\, \sin\, \theta}{\cos\, \theta\, -\, \sin\, \theta} \right )\, \sin\, 2\, \theta\, -\, \displaystyle \frac {1}{2}\, \ln\, (\sec\, 2\,\theta)\, +\, C$$
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