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 :)
 A: 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,
$$\dfrac2{y^2(y^2-2)}=\dfrac{y^2-(y^2-2)}{y^2(y^2-2)}=\dfrac1{y^2-2}-\dfrac1{y^2}$$
A: A hint: $\sin^2(t)+\cos^2 (t)=1$ but
$$
\begin{split}
\sin^4(t)-\cos^4 (t) & =\big(\sin^2(t)-\cos^2 (t)\big)\big(\sin^2(t)+\cos^2 (t)\big)\\
&= \sin^2(t)-\cos^2 (t)
\end{split}
$$
Therefore your integral is
$$
\begin{split}
\int\frac{\sin(t)\cdot\cos^2(t)}{\sin^4(t)-\cos^4(t)}dt=\int\frac{\sin(t)}{\tan^2(t)-1}dt
\end{split}
$$
A: 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}
A: $$
\begin{align}
& \frac 1 2 \int\frac 1 {(x^4-1)\sqrt{x^2+1}} (2x \, dx) \\ {} \\
= {} & \frac 1 2 \int \frac{du}{(u^2-1)\sqrt{u+1}} \\ {} \\
& v = \sqrt{u+1} \\
& v^2 -1 = u \\
& 2v\,dv = du
\end{align}
$$
After that you're integrating a rational function and you may need partial fractions.
