$$\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$$ 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 :)
$$\int\frac{\cos t}{\sin^2 t-\cos^2 t}dt$$
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.
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 $$\int \frac{dt}{(\sinh(t)+1)(\sinh(t)-1)}$$
$$=\frac 12\int (\frac{1}{\sinh(t)-1}-\frac{1}{\sinh(t)+1})dt$$
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}} $
Breaking the $\frac1{x^2-1}$ into partial fractions, the integral becomes
$$\begin{align}\int\frac{\mathrm dx}{(x^2-1)\sqrt{x^2+1}}&=\frac12\int\frac{\mathrm dx}{(x-1)\sqrt{x^2+1}}-\frac12\int\frac{\mathrm dx}{(x+1)\sqrt{x^2+1}}\\&\overset{(1)}{=}-\frac{\text{sgn}(x-1)}{2\sqrt2}\int\frac{\mathrm dt}{\sqrt{t^2+t+\frac12}}+\frac{\text{sgn}(x+1)}{2\sqrt2}\int\frac{\mathrm du}{\sqrt{u^2-u+\frac12}}\\&= -\frac{\text{sgn}(x-1)}{2\sqrt2}\sinh^{-1}\left(t\sqrt2+\frac1{\sqrt2}\right)+\frac{\text{sgn}(x+1)}{2\sqrt2}\sinh^{-1}\left(u\sqrt2-\frac1{\sqrt2}\right)+C\\&= -\frac{\text{sgn}(x-1)}{2\sqrt2}\sinh^{-1}\left(\frac{\sqrt2}{x-1}+\frac1{\sqrt2}\right)+\frac{\text{sgn}(x+1)}{2\sqrt2}\sinh^{-1}\left(\frac{\sqrt2}{x+1}-\frac1{\sqrt2}\right)+C\end{align}$$
Step $(1)$: $t=\frac1{x-1}$ and $u=\frac1{x+1}$
