Question

$\int \frac{e^{x}dx}{(e^x-3e^{-x}+2)}=\frac{1}{k}\log (e^x-1){(e^x+3)}^3.$ Find the value of $k$.

Answer 1

Let $I=\int \frac{e^x}{(e^x-{3e}^{-x}+2)}dx$
Put $t=e^x\Rightarrow dt=e^{x}dx$
$I=\int \frac{1}{t-\frac{3}{t}+2}dt=\int (\frac{3}{4(t+3)}+\frac{1}{4(t-1)})dt$
$=\frac{1}{4}\log (t+3)+\frac{1}{4}\log \left(t-1\right)$
$=\frac{1}{4}\log (e^x+3)+\frac{1}{4}\log (e^x-1)$
$=\frac{1}{4}\log (e^x+3)(e^x-1)$