Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$

Question

Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$

I tried substituting $x=\tan(t)$ in order to get away with square root. ($\:dx=\frac{1}{\cos^2(t)}dt\:$)

$\sqrt{x^2+1}=\frac{1}{\cos(t)}\:\:$ and $\:\:x^4-1=\frac{\sin^4(t)-\cos^4(t)}{\cos^4(t)}$

Now after putting both into main Integral and by simplifying I have :
$$
\int\frac{\sin(t)\cdot\cos^2(t)}{\sin^4(t)-\cos^4(t)} \, dt=\text{?}
$$
Now need a bit help if possible.

Thank you in advance 🙂

Answer 1

Let $\sqrt{x^2+1}=y, dy=\dfrac x{\sqrt{x^2+1}}$

$$\implies x^2=y^2-1$$

$$\int\dfrac{dy}{(y^2-1)^2-1}=\int\dfrac{dy}{y^2(y^2-2)}$$

Now use, $$\dfrac2{y^2(y^2-2)}=\dfrac{y^2-(y^2-2)}{y^2(y^2-2)}=\dfrac1{y^2-2}-\dfrac1{y^2}$$