Question

Evaluate: $\int \frac{dx}{\sqrt{1-e^{2x}}}$

Answer 1

$\int \frac{dx}{\sqrt{1-e^{2x}}}$

$=\int \frac{t\cdot dt}{t(t^2-1)}$

$=\int \frac{dt}{t^2-1}$

$=\frac{1}{2}log\left|\frac{t-1}{t-1}\right|+c$

$=\frac{1}{2}log\left|\frac{\sqrt{1-e^{2x}}-1}{\sqrt{1-e^{2x}}+1}\right|+c$

Let $t=\sqrt{1-e^{2x}}$

$t^2=1-e^{2x}$

$1-t^2=e^{2x}$

$-2tdt=2e^{2x}dx$

$\frac{-t}{e^{2x}}=dx$

$\frac{t}{t^2-1}dt=dx$.