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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integral $\int(2x-3cos\: x+e^x)\;dx$

$\begin{array}{1 1}x^2-3\sin x+e^x+c. \\ x^2+3\sin x+e^x+c.\\ x^2-3\sin x+e^{-x}+c \\ x^2+3\cos x+e^x+c. \end{array} $

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Toolbox:
  • $(i)\;\int x^ndx=\frac{x^{n+1}}{n+1}+c$.
  • $(ii)\;\cos xdx=\sin x+c$.
  • $(ii)\;e^x dx=e^x+c$.
$\int(2x-3\cos x+e^x)dx$.
$\;\;\;=\int 2xdx-3\int \cos x dx+\int e^xdx.$
$\;\;\;=2\int xdx-3\int \cos x dx+\int e^xdx.$
$\;\;\;=2\big(\frac{x^2}{2}\big)-3\sin x+e^x+c.$
$\;\;\;=x^2-3\sin x+e^x+c.$
Hence $\int(2x-3\cos x+e^x)dx=x^2-3\sin x+e^x+c.$
answered Sep 6, 2013 by meena.p
 
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