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

If I=exlog(ex+1)dx, then I equals
  • x+(ex+1)log(ex+1)+C
  • x+(ex+1)log(ex+1)+C
  • x(ex+1)log(ex+1)+C
  • none of these
If I =5x2+xdx, then I equals
  • x+25x+3sin1x+23+C
  • x+25x+3sin1x+27+C
  • x+25x+5sin1x+25+C
  • None of these
cos(lnx)dx=
  • x2(coslnx+sinlnx)
  • x2[coslnxsinlnx]
  • [xcoslnx+sinlnx]
  • 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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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers