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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 3 - MCQExams.com

cosxlog(cosx)dx=sinxlog(cosx)+log|secx+tanx|+f(x)+c, then f(x)=
  • sinx
  • cosx
  • tanx
  • cotx
lf f(x)dx=g(x) then x3f(x2)dx=
  • 12{x2g(x2)g(x2)dx2}
  • 12{x2g(x2)g(x2)dx}
  • 12{x2g(x2)+g(x2)dx2}
  • 12{x2g(x2)+g(x2)dx}
{f(x)g(x)f(x)g(x)}dx=
  • f(x)g(x)g(x)f(x)+c
  • g(x)f(x)f(x)g(x)+c
  • g(x)f(x)+f(x)g(x)+c
  • f(x)g(x)+c
\displaystyle \int (e^\sqrt[3]{x}dx)=
  • x^{2/3}-2 x^{1/3} +2+c
  • (x^{2/3} - 2x^{1/3} +2)\exp(\sqrt[3]{x})+c
  • 3(x^{2/3} - 2x^{1/3}+2)\exp(\sqrt[3]{x})+c
  • 2(x^{2/3} - 2x^{1/3}+2)\exp(\sqrt[3]{x})+c
\displaystyle \int\sqrt{x}.\log xdx=
  • \displaystyle \frac{2}{3}x^{3/2}.\log x-\frac{4}{9}x^{3/2}+c
  • \displaystyle \frac{2}{3}x^{3/2}.\log x+x^{3/2}+c
  • x^{3/2}.(\displaystyle \log x-\frac{2}{3})+c
  • \displaystyle \frac{2}{5}x^{3/2}(\log x+1)+c
lf f(x)dx =g(x) +c, then \displaystyle \int f^{-1}(x)dx is equal to
  • xf^{-1}(x) +c
  • f[g^{-1}(x)]+c
  • x f^{-1}(x)-g[f^{-1}(x)]+c
  • g^{-1}(x) +c
Evaluate \displaystyle \int e^{\sin x}\sin 2xdx
  • 2e^{\sin x}(\sin x+1)+c
  • e^{\sin x}(\sin x+2)+c
  • e^{\sin x}(2\sin x-2)+c
  • e^{\sin x}(3\sin x -2)+c
\displaystyle \int x\tan x\sec^{2}xdx=
  • \displaystyle \frac{1}{2}[x\tan^{2} x-tanx +x]+c
  • \frac{1}{2}[x\tan^{2} x-tanx -x]+c
  • \displaystyle \frac{1}{2}[x\tan^{2}x+ tanx-x]+c
  • x\tan^{2} x-tanx+x+c
\displaystyle \int\frac{2x+sin2x}{1+cos2x}dx=
  • xsin2x+c
  • -xtanx+c
  • xsecx+c
  • xtanx+c
\displaystyle \int e^{x}(3x^{2}+7x+2)dx=
  • e^{x}[3x^{2}+x+1] +c
  • e^{x}[3x^{2}+7x+1] +c
  • e^{x}[3x^{2} - x- l] +c
  • e^{x}[3x^{2}+x]+c

Evaluate \displaystyle \int e^{x}(\log x+\frac{1}{x^{2}})dx
  • e^{x} logx +c
  • \displaystyle e^{x}(\log x-\frac{1}{x})+c
  • \displaystyle e^{x}(\log x+\frac{1}{x})+c
  • \displaystyle \frac{e^{x}}{x^{2}}+c
\displaystyle \int x^{3} \cos x^{2}dx=
  • x^{2} \sin x^{2} -\cos x^{2}+c
  • \displaystyle \frac{1}{2} [ x^{2} \sin x^{2}+ \cos x^{2}] +c
  • \displaystyle \frac{1}{3} [ x^{2} \sin x^{2}+ \cos x^{2}] +c
  • x^{2} \sin x^{2}+ \cos x^{2}+c
\displaystyle \int\frac{\log(x^{2}+a^{2})}{x^{2}}dx =
  • \displaystyle \frac{-\log(x^{2}+a^{2})}{x}+\frac{2}{a}tan^{-1}(\frac{x}{a})+c
  • \displaystyle \frac{\log(x^{2}+a^{2})}{x}+\frac{2}{a}tan^{-1}(\frac{x}{a})+c
  • \displaystyle \frac{\log(x^{2}+a^{2})}{x}-\frac{2}{a}tan^{-1}(\frac{x}{a})+c
  • \displaystyle-\frac{\log(x^{2}+a^{2})}{x}-\frac{2}{a}tan^{-l}(\frac{x}{a})+c
\displaystyle \int\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}dx is equal to
  • x-\sqrt{1-x^{2}}sin^{-1}x+c
  • x+\sqrt{1-x^{2}}sin^{-1}x+c
  • x+sin^{-1}x+c
  • x-sin^{-1}x+c
\displaystyle \int e^{sin^{-1}x}[1+\frac{x}{\sqrt{1-x^{2}}}] dx =
  • xe^{sin^{-1}x}+c
  • e^{sin^{-1}x}+c
  • \displaystyle \frac{1}{\sqrt{1-x^{2}}}e^{\sin^{-1}x}+c
  • x^{2}e^{sin^{1}X}+c
\displaystyle \int e^\sqrt{x}dx=
  • 2 e^\sqrt{x}(\sqrt{x}+1)+c
  • 2 e^ \sqrt{x} \left( \sqrt{x}-1 \right) + c
  • e^ x (x - 1) + c
  • e ^x (x + 1) + c
\displaystyle \int tan^{-1}\sqrt{\frac{1-x}{1+x}}d{x}=
  • \displaystyle \frac{1}{2}[x cos^{-1} x-\sqrt{1-x^{2}}]+c
  • [xcos^{-1}x-\sqrt{1-x^{2}}]+c
  • \displaystyle \frac{1}{2}[x cos^{-1} x+\sqrt{1-x^{2}}]+c
  • x cos^{-1} x+\sqrt{1-x^{2}}+c
(A) : \displaystyle \int(2x tan x \sec^{2}x+\tan^{2}x)dx=x\tan^{2}x+c
(B) : \displaystyle \int[xf^{'}(x)+f(x)]dx=xf(x)+c
  • Both A and R are true and R is the correct explanation of A
  • Both A and R are true but R is not correct explanation of A
  • A is true R is false
  • A is false but R is true.
\displaystyle \int\frac{x-\sin x}{1-\cos x}dx=
  • log |1-Cosx | +c
  • log | x - sin x | +c
  • x\displaystyle \tan\frac{x}{2}+c
  • -x\displaystyle \cot\frac{x}{2}+c
\displaystyle \int(\log x)^{2}dx=
  • x[(\log x)^{2}-2 \log x +2]+c
  • x[(\log x)^{2}+2 \log x +2]+c
  • [(\log x)^{2}-2 \log x +2]+c
  • [(\log x)^{2}+2 \log x +2]+c
\displaystyle \int x^{n}.\log xdx=
  • \displaystyle \frac{x^{n+1}}{(n+1)^{2}}(\log x-\frac{1}{(n+1)^{2}})+c
  • (\displaystyle \log x-\frac{1}{(n+1)^{2}})+c
  • \displaystyle \frac{x^{n+1}}{(n+1)}(\log x-\frac{1}{(n+1)})+c
  • \displaystyle \frac{x^{n+1}}{(n+1)}(\log x-\frac{1}{(n+1)^{2}})+c
\displaystyle \int\frac{\log(1-x)}{x^{2}}dx equal to
  • [(1+\displaystyle \frac{1}{x})\log(1-x)-\log x]+c
  • (1-\displaystyle \frac{1}{x})\log(1-x)-\log x+c
  • (\displaystyle \frac{1}{x}-1)\log(1-x)+\log x+c
  • (\displaystyle \frac{1}{x}-1)\log(1-x)-\log x+c
\int \sin \sqrt {x}dx =
  • 2(Sin\sqrt {x} - \sqrt {x}.\cos \sqrt {x}) + c
  • 2(Sin \sqrt {x} + \sqrt {x}.\cos \sqrt {x}) + c
  • 2 (Sin \sqrt{x} -\dfrac {1}{2} \sqrt {x} .\cos \sqrt {x}) + c
  • 2(Sin \sqrt {x} + \dfrac {1}{2} \sqrt {x} . \cos \sqrt {x}) + c
\displaystyle \int 32x^{3}(\log x)^{2}dx is equal to
  • 8x^{4}(\log x)^{2}+c
  • x^{4}\{8(\log x)^{2}-4\log x+1\}+c
  • x^{4}\{8(\log x)^{2}-4\log x\}+c
  • x^{3}\{(\log x)^{2}+2\log x\}+c
\displaystyle \int x^{3}(\log x)^{2}dx=
  • \displaystyle \frac{x^{4}}{4}[(\log x)^{2}-\frac{\log x}{2}+\frac{1}{8}]+c
  • \displaystyle \frac{x^{4}}{4}[(\log x)^{2}+\frac{\log x}{2}+\frac{1}{8}]+c
  • (\log x)^{2}+2 logx +8+c
  • \displaystyle \frac{x^{4}}{4}[(\log x)^{2}-\frac{\log x}{2}-\frac{1}{8}]+c
\int \displaystyle \frac{sin\frac{x}{2}+cos^{2}\frac{x}{2}}{1+cos x}dx=
  • \sec\frac{x}{2}+x+c
  • \displaystyle \sec\frac{x}{2}+\frac{x}{2}+c
  • \displaystyle\sec\frac{x}{2}-\frac{x}{2}+C
  • \displaystyle\sec\frac{x}{2}-X+C
Solve \displaystyle \int e^{\log x}.\cos xdx
  • x \sin x -cos x + c
  • \displaystyle \frac{x}{2}\sin x+\cos x+c
  • x \sin x + cos x + c
  • x \sin x + cos2 x + c
\displaystyle \int\cos x.\log(\cos x)dx=
  • sin x log (cos x) - log (cos x) + c
  • sin x log (cos x) + sec x + c
  • sin x log (cos x) - sin x + log |sec x + tan x| + c
  • sinx log (cosx) - sec x + c
\displaystyle \int x^{3} sin x^{2}dx=
  • \displaystyle \frac{1}{2}\sin x^{2}-\frac{1}{2}x^{2}\cos x^{2}+c
  • \displaystyle \frac{1}{2}\sin x^{2}+\frac{1}{2}x^{2}\cos x^{2}+c
  • \sin x^{2}-x^{2}\cos x^{2}+c
  • \displaystyle \sin x^{2}-\frac{1}{2}x^{2}\cos x^{2}+c
Evaluate \displaystyle \int x.\sec^{2}x \ dx
  • x \tan x + \log |\sec x| + c
  • x \tan x - \log |\sec x| + c
  • \displaystyle \frac{x}{2}\tan x-\log |\sec x|+c
  • \dfrac{x}{2}\tan x+\log|\sec x|+c
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers