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

Substitue $e ^{ x }= t$ and $e ^{ x } dx = dt$

Therefore

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

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

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

$t ^{2}= u , \quad 2 tdt = du$

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

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

$=\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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