_ ^ e^ m ^ - 1 x 1 + x^2 dx equals to - Vidyadip

MCQ

$\int_{}^{} {\frac{{{e^{m{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $ equals to

A
${e^{{{\tan }^{ - 1}}x}}$
B
$\frac{1}{m}{e^{{{\tan }^{ - 1}}x}}$
✓
$\frac{1}{m}{e^{m{{\tan }^{ - 1}}x}}$
D
None of these

✓

Answer

Correct option: C.

$\frac{1}{m}{e^{m{{\tan }^{ - 1}}x}}$

c
(c) $I = \int {\frac{{{e^{m\,\,{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $,

Put $m{\tan ^{ - 1}}x = t$

==> $\frac{m}{{1 + {x^2}}}\,dx = dt$ ==> $\frac{{dx}}{{1 + {x^2}}} = \frac{{dt}}{m}$

$I = \frac{1}{m}\int {{e^t}.dt} $ $ = \frac{1}{m}{e^t} + c$ $ = \frac{1}{m}{e^{m\,{{\tan }^{ - 1}}x}} + c$.

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