Mathematics

# $\int { \dfrac { { e }^{ x }dx }{ \sqrt { { e }^{ 2x }-1 } } }$

##### SOLUTION

We have,

$\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}$

Let ${{e}^{x}}=t$

On differentiation and we get,

${{e}^{x}}dx=dt$

Then,

$\int{\dfrac{dt}{\sqrt{{{t}^{2}}-1}}}$

$\Rightarrow \log \left| t+\sqrt{{{t}^{2}}-1} \right|+C$

Put $t={{e}^{x}}$

$=\log \left| {{e}^{x}}+\sqrt{{{e}^{x}}-1} \right|+C$

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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