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

Answer 2

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 $$ \begin{split} \int\frac{\sin(t)\cdot\cos^2(t)}{\sin^4(t)-\cos^4(t)}dt=\int\frac{\sin(t)}{\tan^2(t)-1}dt \end{split} $$

Answer 3

$$ \begin{align} & \frac 1 2 \int\frac 1 {(x^4-1)\sqrt{x^2+1}} (2x \, dx) \\ {} \\ = {} & \frac 1 2 \int \frac{du}{(u^2-1)\sqrt{u+1}} \\ {} \\ & v = \sqrt{u+1} \\ & v^2 -1 = u \\ & 2v\,dv = du \end{align} $$ After that you're integrating a rational function and you may need partial fractions.

Answer 4

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}