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

Integrate : $\int \large\frac{\tan^{-1} x}{1+x^2}$$ dx$

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

1 Answer

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$\tan ^{-1}=t$
$1/1+x^2.dx=dt$
$\int t.dt$
$\qquad=\large\frac {t^2}{2}$$+c$
$ \qquad=\large\frac{(\tan ^{-1}x)^2}{2}$$+c $
Hence b is the correct answer.
answered Dec 19, 2013 by meena.p
 
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