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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