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

Evaluate xa2x2a2+x2dx
  • 12a2cos1(x2a2)+12a4+x4+C
  • 12sin1(x2a2)+a4+x4+C
  • 12a2sin1(x2a2)+12a4x4+C
  • 12cos1(x2a2)+12a4x4+C
Solve:
4x2e2x3 dx
  • 23e2x3+C
  • 43e2x3+C
  • 32e2x3+C
  • 13e2x3+C

The value of lnn(1(1x))dxx(x1) is

  • 12[m(11x)2]+C
  • 12[m(1+1x)]2+C
  • 12n(x(x1))+C
  • 12[n(x(x1))]2+C
If I=ex(1x)2(1+x2)2dx , then I =
  • ex1x2(1+x2)+C
  • ex1x(1+x2)+C
  • ex1(1+x2)+C
  • ex12x(1+x2)2+C
The value of etan1x(1+x+x21+x2)dx is equal to
  • xetan1x+C
  • x2etan1x+C
  • 1xetan1x+C
  • xecot1x+C
Evaluate xx2+2dx
  • I=x22+C
  • I=x2+2+C
  • I=x3+2+C
  • I=x32+C
The function f(x) satisfying the equation f2(x)+4f(x).f(x)+[f(x)]2=0 is-
  • f(x)=c.e(23x)
  • f(x)=c.e(23x)
  • f(x)=c.e(32)x
  • f(x)=c.e(2+3)x
The value of (x1)ex is equal to 
  • xex+C
  • xex+C
  • xex+C
  • xex+C
(1+3x+3x2+4x3+........)dx(|x|<1)-
  • (1+x)1+c
  • (1x)1+c
  • (1+x)2+c
  • None of these
2x14xdx=Ksin1(2x)+C, then the value of K is equal to
  • n2
  • 122
  • 12
  • 1n2
cos(logx)dx=............+c
  • x2[cos(logx)+sin(logx)]
  • x4[cos(logx)+sin(logx)]
  • x2[cos(logx)sin(logx)]
  • x2[sin(logx)+cos(logx)]
xsinxsec3xdx=
  • 12[sec2xtanx]+c
  • 12[xsec2xtanx]+c
  • 12[xsec2x+tanx]+c
  • 12[sec2x+tanx]+c
P(x)ekxdx=Q(x)e4x+C, where P(x) is polynomial of degree n and Q(x) is polynomial of degree 7. Then the value of n+7+k+limxP(x)Q(x) is:
  • 18
  • 19
  • 20
  • 22
Evaluate using limit of sum:
31(x+1)2dx
  • 26
  • 28
  • 30
  • 32
π/40sec2x(1+tanx)(2+tanx)dx equals:
  • loge23
  • loge3
  • 12loge43
  • loge43
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}
\int {{e^x}\left[ {{\mathop{\rm tanx}\nolimits}  - log\left( {\cos x} \right)} \right]} dx =
  • {e^x}\log \left( {\sec x} \right) + c
  • {e^x}\log \left( {co\sec x} \right) + c
  • {e^x}\log \left( {\cos x} \right) + c
  • {e^x}\log \left( {\sin x} \right) + c
If \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 :
  • -2 & \dfrac{2}{3}
  • 2 & -\dfrac{2}{3}
  • 2 & \dfrac{2}{3}
  • None of these
\displaystyle \int (x^2-x+5)\, dx
  • \dfrac {x^3}3-\dfrac{x^2}2+5x+c
  • \dfrac {x^3}3+\dfrac{x^2}2+5x+c
  • \dfrac {x^2}2-\dfrac{x}2+5x+c
  • \dfrac {x^4}4-\dfrac{x^4}3+5
\displaystyle \int {{x^3}{e^{{x^2}}}dx = }
  • \dfrac{1}{2}\left( {{x^2} + 1} \right){e^{{x^2}}} + c
  • \left( {{x^2} + 1} \right){e^{{x^2}}} + c
  • \dfrac{1}{2}\left( {{x^2} - 1} \right){e^{{x^2}}} + c
  • \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?
  • \sin\left(e^{x^3}\right)+C
  • 3\sin \left(e^{x^3}\right)+C
  • \dfrac{1}{3}\sin\left(e^{x^3}\right)+C
  • e^x\sin\left(e^{x^3}\right)+C
The value of \int  \sqrt { \dfrac { e^ x\quad -\quad 1 }{ e^ x\quad +\quad 1 }  } dx is equal to
  • \ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) - sec^ -1(e^ x)+c
  • \ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) +sec^ -1(e^ x)+c
  • \ell n\left( e^ x-\sqrt { e^ 2x\quad -1 } \right) -sec^ -1(e^ x)+c
  • \ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) -sin^ -1(e^ {-x})+c
\int \sqrt {1 + \sin x}dx =
  • \dfrac {1}{2}\left (\sin \dfrac {x}{2} + \cos \dfrac {x}{2}\right ) + c
  • \dfrac {1}{2}\left (\sin \dfrac {x}{2} - \cos \dfrac {x}{2}\right ) + c
  • 2\sqrt {1 + \sin x} + c
  • -2\sqrt {1 - \sin x} + c
The integral of \displaystyle\int e^{\sin x}(x\cos x-\sec x\tan x)dx is?
  • se^{\sin x}-e^{\sin x}\sec x+c
  • (x+\sec x)e^{\sin x}+c
  • e^{\sin x}\cos x+c
  • e^{\sin x}(\cos x-\sec x)+c
The value of \int (x-1) e^{-x} dx is equal to
  • -xe^x + C
  • xe^x + C
  • -xe^{-x} + C
  • xe^{-x} + C
\int \dfrac {dx}{(x^{2} + 4x + 5)^{2}} is equal to
  • \dfrac {1}{2}\left [\tan^{-1}(x + 1) + \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c
  • \dfrac {1}{2}\left [\tan^{-1}(x + 2) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c
  • \dfrac {1}{2}\left [\tan^{-1}(x + 1) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c
  • \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=
  • (e^x+e^{-x}) tan (x/2)+C
  • (e^x-e^{-x}) cot (x/2)+C
  • (e^x-e^{-x}) tan (x/2)+C
  • (e^x-e^{-x}) cosec (x/2)+C
If \displaystyle \int \dfrac{x^4 + 1}{x^6 + 1} dx = \tan^{-1} (f(x)) -\dfrac{2}{3} \tan^{-1} (g(x)) + C, then
  • Both f(x) & g(x) are odd functions
  • g(x) is monotonic function
  • none of these
  • None 
Integral of f ( x ) = \sqrt { 1 + x ^ { 2 } } with respect to x ^ { 2 } is

  • \frac { 2 } { 3 } \frac { \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } } { x } + k
  • \frac { 2 } { 3 } \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k
  • \frac { 2 } { 3 } x \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k
  • None of these
 The value of \displaystyle \int {\dfrac{{dx}}{{\sin x.\sin \left( {x + \alpha } \right)}}} equal to
  • \csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sin x}}{{\sin \left( {x + \alpha } \right)}}} \right| + c
  • \;\;\;\;\csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sin \left( {x + \alpha } \right)}}{{\sin x}}} \right| + c
  • \csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sec \left( {x + \alpha } \right)}}{{\sec x}}} \right| + c\;
  • \,\;\;\csc\,\,\alpha \,\,\ell n\left| {\dfrac{{\sec x}}{{\sec \left( {x + \alpha } \right)}}} \right| + c
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