Answer
$$\int \frac{1}{e^{2\theta}-1}d\theta=\sec^{-1}(e^\theta)+C$$
Work Step by Step
$$A=\int \frac{1}{e^{2\theta}-1}d\theta=\int \frac{1}{(e^{\theta})^2-1^2}d\theta$$
We set $u=e^\theta$.
Then $$du=e^\theta d\theta=u d\theta$$ $$d\theta=\frac{du}{u}$$
Therefore, $$A=\int\frac{du}{u(u^2-1^2)}$$
We have $$\int\frac{dx}{x(x^2-a^2)}=\frac{1}{a}\sec^{-1}\Big(\frac{x}{a}\Big)+C$$
So $$A=\frac{1}{1}\sec^{-1}\Big(\frac{u}{1}\Big)+C$$ $$A=\sec^{-1}u+C$$ $$A=\sec^{-1}(e^\theta)+C$$
