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

Integrate: $\int \large\frac{\Large e^{\tan ^{-1}(m)}}{1+m^2}$$.dm$

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

1 Answer

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