# Evaluating $\int\frac{1}{\sqrt {e^{4x}-1}} dx$ using the expression $\frac{1}{a}\tan^{-1}\left(\frac{1}{a}\right)+c$

Given

$$\int\frac{1}{\sqrt {e^{4x}-1}} dx ,$$

is it possible to find the integral by $u$-substitution and using the formula: $\frac{1}{a}\tan^{-1}\left(\frac{1}{a}\right)+c$. If so, how?

• If you're looking to use trigonometric substitution, the expression of the form $\sqrt{(e^{2 x})^2 - 1}$ in the denominator suggests substituting $u = e^{2x}$, so that $du = 2 e^{2x} dx = 2 u dx$ and hence $dx = 2 \frac{du}{u}$. The obvious substitution after this point does not lead to the particular expression you mention. (Also, the "formula" you wrote is not a formula, but rather an expression.) Feb 27, 2016 at 18:05

Let $\displaystyle u=\frac{1}{\sqrt{e^{4x}-1}},\;\;$ so $\displaystyle du=\frac{-2e^{4x}}{(e^{4x}-1)^{3/2}}dx$.
Then $\displaystyle\int\frac{1}{\sqrt{e^{4x}-1}}dx=-\frac{1}{2}\int\frac{e^{4x}-1}{e^{4x}}\cdot\frac{-2e^{4x}}{(e^{4x}-1)^{3/2}}dx=-\frac{1}{2}\int\frac{1}{\frac{1}{e^{4x}-1}+1}\cdot\frac{-2e^{4x}}{(e^{4x}-1)^{3/2}}dx$
$\displaystyle=-\frac{1}{2}\int\frac{1}{u^2+1}du=-\frac{1}{2}\tan^{-1}u+C=-\frac{1}{2}\tan^{-1}\left(\frac{1}{\sqrt{e^{4x}-1}}\right)+C$
Also, $\int \dfrac{1}{\sqrt{e^{4x}-1}} dx$=$\int \dfrac{1}{\sqrt{e^{4x}-1}} \dfrac{e^x}{e^x}$. Making the substitution $u=e^x$; $du=e^x dx$, we get
$\int \dfrac{1}{u\sqrt{u^4-1}} du$. Now, if we put $u=\sqrt {sec t}$, $du=\dfrac{sec t \cdot tg t}{2\sqrt{sec t}}$ we get $\dfrac{1}{2}\int 1 dt=\dfrac{t}{2}$ and turning back.. $\int \dfrac{1}{\sqrt{e^{4x}-1}} dx=\dfrac{sec^{-1}(e^{2x})}{2}$ This method usually works for functions involving only e^x.