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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \large\frac{\tan ^{-1} (e^x)}{1+(e^x)^2}.e^x\;dx$

\[\begin {array} {1 1} (a)\;\tan ^{-1}(e^x)+c \\ (b)\;[\tan ^{-1}(e^x)]^2+c \\ (c)\;\frac{[\tan ^{-1}(e^x)]^2}{2}+c \\ (d)\;None \end {array}\]

1 Answer

$\tan ^{-1} (e^x)=d$
$\large\frac{e^x}{1+(e^x)^2}.$$dx=dt$
=> $ \int t\;dt$
=> $\large\frac{t^2}{2} $$+c$
$ \large\frac{[\tan ^{-1}(e^x)]^2}{2}+c$
Hence c is the correct answer.
answered Dec 19, 2013 by meena.p
 
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