Solve integral $\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$

Question

$$\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$$ I'm trying to solve this integral. First I substituted : $x=\tan(t)$; $t=\arctan(x)$

Then $$ dx=\frac{1}{\cos^2(t)}\,dt$$

Now by simplifying I'm to this step $$ \int\frac{\cos(t)}{\sin^2(t)-\cos^2(t)}\,dt$$

What can i do now ..

Thank you in advance :)

Answer 1

$$\int\frac{\cos t}{\sin^2 t-\cos^2 t}dt$$

Use $\cos^2 t = 1-\sin^2 t$ to modify the denominator $$\int\frac{\cos t}{2\sin^2 t-1}dt$$ Substitute $u=\sin t$, $du =\cos t \, dt$ $$\int\frac{1}{2u^2-1}du$$ Now integrate using partial fractions or otherwise.

Answer 2

Substitute $t= \frac{\sqrt2 x}{\sqrt{x^2+1}}$

\begin{align} I=& \int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx\\ = &\frac1{\sqrt2} \int \frac{1}{t^2-1} dt=\frac1{2\sqrt2} \ln|\frac{t-1}{t+1}| = \frac1{2\sqrt2} \ln|\frac{\sqrt2 x-\sqrt{x^2+1}}{\sqrt2 x+\sqrt{x^2+1}}| \end{align} Alternatively $I=-\frac1{2\sqrt2}\coth^{-1}\frac{3x+\frac1x}{2\sqrt2 \sqrt{x^2+1}} $

Answer 3

Other approach

Put $ x=\sinh(t) $ with

$$dx=\cosh(t)dt$$

$$\sqrt{x^2+1}=\sqrt{\sinh^2(t)+1}=\cosh(t)$$

$$x^2-1=(\sinh^2(t)-1)$$

It becomes $$\int \frac{dt}{(\sinh(t)+1)(\sinh(t)-1)}$$

$$=\frac 12\int (\frac{1}{\sinh(t)-1}-\frac{1}{\sinh(t)+1})dt$$