, , ^ - 1 ( e^x ) e^x dx is equal to : - Vidyadip

MCQ

$\int {\,\,\frac{{{{\cot }^{ - 1}}({e^x})}}{{{e^x}}}} $ $dx $ is equal to :

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

✓

Answer

Correct option: C.

$\frac{1}{2}$ $ln$ $ (e^{2x} + 1) - $$\frac{{{{\cot }^{ - 1}}({e^x})}}{{{e^x}}}$ $ - x + c$

c
Let $\int \frac{\cot ^{-1}\left(e^{x}\right)}{e^{x}} d x=I$

Substitue $e^{x}=t$ and $e^{x} d x=d t$

Therefore

$I=\int \frac{\cot ^{-1}(t)}{t^{2}} d t$

Integrating it by part, taking $\cot ^{-1}(t)$ as first function and $t^{2}$ as second function, we get $I=-\frac{\cot ^{-1}(t)}{t}-\int \frac{1}{t\left(t^{2}+1\right)} d t$

$\int \frac{1}{t\left(t^{2}+1\right)} d t=P$

$t^{2}=u, \quad 2 t d t=d u$

$P=\frac{1}{2} \int \frac{1}{u(u+1)} d u$

$=\frac{1}{2} \int\left(\frac{1}{u}-\frac{1}{(u+1)}\right) d u$

$=\frac{1}{2}(\log u-\log u+1)+C$

$I=-\frac{\cot ^{-1}(t)}{t}-\frac{1}{2}(\log |u|-\log |u+1|)+C$

$=\frac{1}{2}\left(\log \left|t^{2}+1\right|-\log \left|t^{2}\right|-\frac{2 \cot ^{-1}(t)}{t}\right)+C$

$=\frac{1}{2}\left(\log \left|e^{2 x}+1\right|-\log \left|e^{2 x}\right|-\frac{2 \cot ^{-1}\left(e^{x}\right)}{e^{x}}\right)+C$

$=\frac{1}{2} \log \left|e^{2 x}+1\right|-\frac{\cot ^{-1}\left(e^{x}\right)}{e^{x}}-x+C$

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