Question

Solve :
$\int \frac{dy}{e^{(x-1)y}-1}$

Answer 1

Consider, $I=\int \frac{dy}{e^{(x-1)y}-1}$

$\Rightarrow I=\int \frac{dy}{\frac{e^{xy}}{e^y}-1}$

$=\int \frac{e^{y}dy}{e^{xy}-e^y}$

$e^y=t$ $\Rightarrow e^{y}dy=dt$

$I=\int \frac{dt}{e^{x}t-t}$

$=\int \frac{dt}{t(e^x-1)}$

$=\frac{1}{(e^x-1)}\int \frac{dt}{t}$

$=\frac{1}{(e^x-1)}(n\left|t\right|+\left|n\right|c)$

$=c{t}^{(e^x-1)}$

$=c{\left(e^y\right)}^{e^x-1}$

$\Rightarrow c{e}^{y(e^x-1)}$