_ ^ dx e^x + 1 - 2 e^ - x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{{e^x} + 1 - 2{e^{ - x}}}} = } $

A
$\log ({e^x} - 1) - \log ({e^x} + 2) + c$
B
$\frac{1}{2}\log ({e^x} - 1) - \frac{1}{3}\log ({e^x} + 2) + c$
✓
$\frac{1}{3}\log ({e^x} - 1) - \frac{1}{3}\log ({e^x} + 2) + c$
D
$\frac{1}{3}\log ({e^x} - 1) + \frac{1}{3}\log ({e^x} + 2) + c$

✓

Answer

Correct option: C.

$\frac{1}{3}\log ({e^x} - 1) - \frac{1}{3}\log ({e^x} + 2) + c$

c
(c)$\int_{}^{} {\frac{{{e^x}dx}}{{{e^{2x}} + {e^x} - 2}}} = \int_{}^{} {\frac{{dt}}{{{t^2} + t - 2}}} $ $\left\{ {\,\,\,{e^x} = t \Rightarrow {e^x}dx = dt} \right\}$
$ = \int_{}^{} {\frac{{dt}}{{(t + 2)(t - 1)}}} = \,\int_{}^{} {\frac{1}{3}\left[ {\frac{1}{{t - 1}} - \frac{1}{{t + 2}}} \right]} \,dt$
$ = \frac{1}{3}\log ({e^x} - 1) - \frac{1}{3}\log ({e^x} + 2) + c.$

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