CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 13 - MCQExams.com

The value of $$\displaystyle \int e^{x} \left [\dfrac {1 + \sin x}{1 + \cos x}\right ] dx$$ is
  • $$\dfrac {1}{2} e^{x} \sec \dfrac {x}{2} + C$$
  • $$e^{x} \sec \dfrac {x}{2} + C$$
  • $$\dfrac {1}{2} e^{x} \tan \dfrac {x}{2} + C$$
  • $$e^{x} \tan \dfrac {x}{2} + C$$
$$\displaystyle\int { \cfrac { { x }^{ 2 }-1 }{ { x }^{ 4 }+{ x }^{ 2 }+1 }  } dx$$ is equal to
  • $$\log { \left( { x }^{ 4 }+{ x }^{ 2 }+1 \right) } +c$$
  • $$\log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c } $$
  • $$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c } $$
  • $$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 } +c } $$
$$\displaystyle \int\frac{dx}{x(x^7+1)}$$ is equal to 
  • $$log\left(\dfrac{x^7}{x^7+1}\right)+c$$
  • $$\dfrac{1}{7}log\left(\dfrac{x^7}{x^7+1}\right)+c$$
  • $$log\left(\dfrac{x^7+1}{x^7}\right)+c$$
  • $$\dfrac{1}{7}log\left(\dfrac{x^7+1}{x^7}\right)+c$$
$$\displaystyle \int {\frac{x^2}{(x\,\sin\,x+\cos\,x)^2}} dx$$ is equal to
  • $$\displaystyle \frac{\sin\,x+\cos\,x}{x\,\sin\,x+\cos\,x}+C$$
  • $$\displaystyle \frac{x\,\sin\,x-\cos\,x}{x\,\sin\,x+\cos\,x}+C$$
  • $$\displaystyle \frac{\sin\,x-x\,\cos\,x}{x\,\sin\,x+\cos\,x}+C$$
  • None of these
$$\displaystyle \int \displaystyle \frac{x\, -\, \sin x}{1\, +\, \cos x}\,dx\, =\, x \tan \left ( \displaystyle \frac{x}{2} \right )\, +\, p\, \log \left | \sec \left ( \displaystyle \frac{x}{2} \right ) \right |\, +\, c\, \Rightarrow\, p\, =$$
  • -4
  • 4
  • 2
  • -2
Let $$f(x)\, =\, 3x^{2}.\, \sin \,\displaystyle \frac{1}{x}\, -\, x\cos\, \displaystyle \frac{1}{x},\, x\, \neq\, 0, f(0)\, =\, 0\, f \left (\, \displaystyle \frac{1}{\pi} \right )\, =\, 0$$, then which of the  following is/are not correct.
  • $$f(x)$$ is continuous at $$x = 0$$
  • $$f(x)$$ is non-differentiable at $$x = 0$$
  • $$f(x)$$ is discontinuous at $$x = 0$$
  • $$f(x)$$ is differentiable at $$x = 0$$
$$\displaystyle\int { \cfrac { \sqrt { x }  }{ \sqrt { x } -\sqrt [ 3 ]{ x }  }  } dx$$ is equal to
  • $$6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 2 } +\cfrac { { x }^{ 1/3 } }{ 3 } +\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$$
  • $$6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 3 } +\cfrac { { x }^{ 1/3 } }{ 2 } +\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$$
  • $$6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 2 } +\cfrac { { x }^{ 1/3 } }{ 3 } +{ x }^{ 1/6 }+\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$$
  • None of the above
Evaluate $$\displaystyle {\int \sin^{-1}\, \sqrt {\frac {x}{a\, +\, x}} dx}$$
  • $$(a + x) \tan^{-1} \displaystyle \sqrt \frac {x}{a}\, -\, \sqrt {ax}\, +\, c$$
  • $$\displaystyle \left(\frac { ax+1 }{ 2a } \right)\sin^{ -1 }\, \sqrt { \frac { x }{ a\, +\, x } } -\frac { 1 }{ 2 } \sqrt { \frac { x }{ a } } +c$$
  • $$a \tan^{-1} \displaystyle \sqrt \frac {x}{a}\, -\, \sqrt {ax}\, +\, c$$
  • $$(a + x) \tan^{-1} \displaystyle \sqrt {x}\, -\, \sqrt {ax}\, +\, c$$
$$\displaystyle \int x \frac{\ln (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}\, dx$$ equals to
  • $$\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) - x + c$$
  • $$\displaystyle \frac{x}{2} \ln^{2} (x + \sqrt{1 + x^{2}}) - \frac{x}{\sqrt{1 + x^{2}}} + c$$
  • $$\displaystyle \frac{x}{2} \ln^{2} (x + \sqrt{1 + x^{2}}) + \frac{x}{\sqrt{1 + x^{2}}} + c$$
  • $$\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) + x + c$$
$$\displaystyle \int { \cfrac { { x }^{ 3 } }{ \sqrt { 1+x^2 }  }  } dx$$
  • $$\sqrt { 1+x } -\cfrac { x }{ 3 } { \left( 1+{ x }^{ 2 } \right) }^{ 3/2 }+c$$
  • $$x\sqrt { 1+{ x }^{ 2 } } +\cfrac { 2 }{ 3 } { \left( 1+{ x }^{ 2 } \right) }^{ 3/2 }+c$$
  • $$\dfrac{{ x }^{ 2 }\sqrt { 1+{ x }^{ 2 } }}{3}-\cfrac { 2 }{ 3 } {\sqrt{1+{ x }^{ 2 }} }+c$$
  • $${ x }^{ 2 }\sqrt { 1+{ x }^{ 2 } } -\cfrac { 1 }{ 3 } { \left( 1+{ x }^{ 2 } \right) }^{ 3/2 }+c$$
$$\displaystyle \int { \sqrt { x } { e }^{ \sqrt { x }  } } dx$$ is equal to:
  • $$\displaystyle 2\sqrt { x } -{ e }^{ \sqrt { x } }-4\sqrt { { xe }^{ \sqrt { x } } } +c$$
  • $$\displaystyle \left( 2x-4\sqrt { x } +4 \right) { e }^{ \sqrt { x } }+c$$
  • $$\displaystyle \left( 2x+4\sqrt { x } +4 \right) { e }^{ \sqrt { x } }+c$$
  • $$\displaystyle \left( 1-4\sqrt { x } \right) { e }^{ \sqrt { x } }+c$$
$$\displaystyle\int \dfrac{cos^{n-1}}{sin^{n+1}}dx, n\neq 0$$ is ____
  • $$\dfrac{cot^nx}{n}$$
  • $$\cfrac{-cot^{n-1}x}{n-1}$$
  • $$\cfrac{-cot^nx}{n}$$
  • $$\cfrac{cot^{n-1}x}{n-1}$$
$$\int { { ({ x }^{ 2 }+5) }^{ 3 } } dx$$
  • $$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$
  • $$\cfrac { { x }^{ 6 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$
  • $$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }-25{ x }^{ 3 }+125x+C$$
  • $$\cfrac { { x }^{ 7 } }{ 7 } +4{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$
Evaluate: $$\displaystyle \int \dfrac{(x-1)e^x}{(x+1)^3}dx=$$
  • $$\dfrac{e^x}{x+1}$$
  • $$\dfrac{e^x}{(x+1)^2}$$
  • $$\dfrac{e^x}{(x+1)^3}$$
  • $$\dfrac{x\cdot e^x}{(x+1)}$$
$$\int { { x }^{ 4 }{ e }^{ 2x } } dx=$$
  • $$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$$
  • $$\cfrac { { e }^{ 2x } }{ 2 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$$
  • $$\cfrac { { e }^{ 2x } }{ 8 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$$
  • $$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$$
If $$\displaystyle \int \dfrac{\text{cosec}^2x - 2010}{\cos^{2010x}}dx=-\dfrac{f(x)}{(g(x))^{2010}}+C$$; then the number of solutions where equation $$\dfrac{f(x)}{g(x)}=\left \{ x \right \}$$ in $$[0, 2\pi]$$ is/are:
  • $$0$$
  • $$1$$
  • $$2$$
  • $$3$$
Let $$f(x)$$ be a positive function. Let
$$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$$,
$$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$$,
where $$2k - 1 > 0$$, then $$\dfrac {I_{1}}{I_{2}}$$ is
  • $$2$$
  • $$k$$
  • $$\dfrac {1}{2}$$
  • $$1$$
If $$\displaystyle \int \dfrac{x + \sqrt[3]{x^2} + \sqrt[6]{x}}{x(1+\sqrt[3]{x})} dx = ax^{\frac{2}{3}} + b\ \tan^{-1}(\sqrt[6]{x}) +c $$ then:
  • $$ a= \frac{3}{2}$$
  • $$ a= \frac{3}{4}$$
  • $$ a= \frac{-3}{2}$$
  • $$ a= \frac{-3}{4}$$
Let $$y = y(x), y(1)=1\ and\ y(e) ={e^2}$$ . Consider
$$J = \int {{{x + y} \over {xy}}} dy$$, $$I = \int {{{x + y} \over {{x^2}}}} dx$$, $$J - I = g\left( x \right)$$ and g(1) = 1, then the value of g(e) is
  • $$2e+1$$
  • $$e+1$$
  • $${e^2}$$-e+1
  • $${e^2}$$+e-1
If $$\int { f(x)dx=g(x) } $$, then $$\int { { f }^{ -1 }(x)dx } $$ is 
  • $$x{ f }^{ -1 }(x)+C$$
  • $$f({ g }^{ -1 }(x))+C$$
  • $$x{ f }^{ -1 }(x)-g({ f }^{ -1 }(x))+C$$
  • $${ g }^{ -1 }(x)+C$$
The value of $$\displaystyle \int{ \sqrt{\dfrac{e^x - 1}{e^x + 1}}dx}$$ is equal to
  • $$ln (e^x + \sqrt{e^{2x} - 1} - sec^{-1} (e^x) + C$$
  • $$ln (e^x + \sqrt{e^{2x} - 1} + sin^{-1} (e^x) + C$$
  • $$ln ( e^x - \sqrt{e^{2x} - 1} - sec^{-1} (e^x) + C$$
  • none of these
Evaluate :
$$\int { \dfrac { { x }^{ 2 }\left( x\sec ^{ 2 }{ x } -\tan { x }  \right)  }{ \left( x\tan { x } -1 \right) ^{ 2 } }  }dx =$$ ?
  • $$2\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x+1}+C$$
  • $$x\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x+1}+C$$
  • $$2\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x-1}+C$$
  • none
If $$\displaystyle\int { { x }^{ 13/2 }.{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 1/2 }dx } =P{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 7/2 }+Q{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 5/2 }+R{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 3/2 }+C$$, then $$P,\ Q$$ and $$R$$ are
  • $$P=\frac { 4 }{ 35 } ,\ Q=-\frac { 8 }{ 25 } ,\ R=\frac { 4 }{ 15 } $$
  • $$P=\frac { 4 }{ 35 } ,\ Q=\frac { 8 }{ 25 } ,\ R=\frac { 4 }{ 15 } $$
  • $$P=-\frac { 4 }{ 35 } ,\ Q=-\frac { 8 }{ 25 } ,\ R=\frac { 4 }{ 15 } $$
  • $$P=\frac { 4 }{ 35 } ,\ Q=-\frac { 8 }{ 25 } ,\ R=-\frac { 4 }{ 15 } $$
The value of $$\int (x\ e^{\ell n \sin x}-\cos x)dx$$ is equal to:
  • $$x \cos x+C$$
  • $$\sin x-x\ \cos +C$$
  • $$-e^{\ell n x}\cos x+C$$
  • $$\sin x+x \cos x+C$$
One of the roots of the equation $$2000x^6+100x^5+10x^3+x-2=0$$ is of the form $$\dfrac{m+\sqrt{n}}{r}$$. When 'm' is non zero integer and n and r relatively prime natural numbers. Then $$\dfrac{m+n+r}{100}=?$$
  • $$100$$
  • $$2$$
  • $$3$$
  • $$0$$
$$\int {x{{\sin }^{ - 1}}xdx} $$=?
  • $$\frac{1}{4}{\sin ^{ - 1}}(x)(2x + 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$$
  • $$\frac{1}{4}{\sin ^{ - 1}}(x)(2x^2 - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$$
  • $$\frac{1}{4}{\cos ^{ - 1}}(x)(2x - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$$
  • $$\frac{1}{4}{\sin ^{ - 1}}(x)(2x - 1) - \frac{{x\sqrt {1 - {x^2}} }}{4} + c$$
Solve : $$\int x . \sin^2 x \, dx$$
  • $$\dfrac {x^2}{4}+x\dfrac {\sin 2x}{6}+\dfrac {\cos 2x}{8}$$
  • $$\dfrac {x^2}{4}-x\dfrac {\sin 2x}{4}-\dfrac {\cos 2x}{8}$$
  • $$\dfrac {x^2}{6}+x\dfrac {\sin 2x}{4}-\dfrac {\cos 2x}{8}$$
  • $$\dfrac {x^2}{4}-x\dfrac {\cos 2x}{4}-\dfrac {\sin  2x}{8}$$
The value of $$\displaystyle\int\left(\dfrac{x^2+\cos^2x}{1+x^2}\right)cosec^2xdx$$ equals?
  • $$\cot x+\tan^{-1}x+c$$
  • $$\cot x-\tan^{-1}x+c$$
  • $$-(\cot x+\tan^{-1}x)+c$$
  • $$-\cot x-\cot^{-1}x+c$$
The value of the integer $$I=\int_{1}^{\infty} \dfrac{(x^{2}-x)}{x^{3}\sqrt{(x^{2}-1)}}dx$$ is
  • $$0$$
  • $$2/3$$
  • $$4/3$$
  • $$none\ of\ these$$
The integral $$\int _{ \tfrac { \pi  }{ 12 }  }^{ \tfrac { \pi  }{ 4 }  }{ \dfrac { 8\cos { 2x }  }{ { \left( \tan { x } +\cot { x }  \right)  }^{ 3 } } dx }$$ equals:
  • $$\dfrac {15}{128}$$
  • $$\dfrac {5}{64}$$
  • $$\dfrac {13}{32}$$
  • $$\dfrac {13}{256}$$
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