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

Note
\begin{align}
\int\frac{\sin t\cos^2t}{\sin^4t-\cos^4t} \, dt
= &\int\frac{\sin t\cos^2t}{(1-\cos^2t)^2-\cos^4t} \, dt
=\int\frac{\sin t\cos^2t}{1-2\cos^2t} \, dt\\
 = & \frac12 \int\left( 1- \frac{1}{1-2\cos^2t} \right) \, d(\cos t)\\
=&\frac12\cos t -\frac1{4\sqrt2}\ln \frac{1+\sqrt2\cos t}{1-\sqrt2\cos t}
\end{align}
