Question

integrtion of 1/e^x+be^-x

Answer 1

Step-by-step explanation:

We have,

$\int \frac{1}{ {e}^{x} + b {e}^{ - x} } dx$

$= \int \frac{ {e}^{x}dx }{ {e}^{2x} + b }$

$let \: \: {e}^{x} = t \\ {e}^{x} dx = dt$

$= \int \frac{dt}{ {t}^{2} + b }$

$= \int \frac{dt}{ {t}^{2} + ({ \sqrt{b} })^{2} }$

$= \frac{1}{ \sqrt{b} } \tan^{ - 1} ( \frac{t}{ \sqrt{b} } ) + c$

$= \frac{1}{ \sqrt{b} } \tan^{ - 1} ( \frac{ {e}^{x} }{ \sqrt{b} } ) + c$