Question

dx/e*x-e*-x ka hole square


​

Answer 1

The correct question is

Integrate : $\int \frac{dx}{(e^{x}-e^{-x})^{2}}$

Solution :

$\textsf{Now,}\: \int \frac{dx}{(e^{x}-e^{-x})^{2}}$

$= \int \frac{e^{2x}dx}{(e^{2x}-1)^{2}}$

$= \frac{1}{2} \int \dfrac{dz}{z^{2}}$

{ Taking e²ˣ - 1 = z

⇒ 2 e²ˣ dx = dz

⇒ e²ˣ dx = $\frac{1}{2}$ dz }

$= - \frac{1}{2z} \: +c$ ,

where c is integral constant

$= - \frac{1}{2 (e^{2x}-1)}+c$ ,

which is the required integral.