1 + ^2 x 1 - ^2 x ,dx equals to - Vidyadip

MCQ

$\int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}\,dx} $ equals to

A
$\log \left( {\frac{{1 - \tan x}}{{1 + \tan x}}} \right) + c$
B
$\log \left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right) + c$
C
$\frac{1}{2}\log \left( {\frac{{1 - \tan x}}{{1 + \tan x}}} \right) + c$
✓
$\frac{1}{2}\log \left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right) + c$

✓

Answer

Correct option: D.

$\frac{1}{2}\log \left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right) + c$

d
(d) $I = \int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}dx} $$ = \int {\frac{{{{\sec }^2}x}}{{1 - {{\tan }^2}x}}dx} $
Put $\tan x = t$ ==> ${\sec ^2}x.\,dx = dt$ ==> $I = \int {\frac{{dt}}{{1 - {t^2}}}} $
$ = \frac{1}{{2 \times 1}}\log \left[ {\frac{{1 + t}}{{1 - t}}} \right] + c$$ = \frac{1}{2}\log \left| {\frac{{1 + \tan x}}{{1 - \tan x}}} \right| + c$.

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