How can I do this integral? How can I compute this integral?
$$\int\frac{x}{\sqrt{1+x^2}-x^2-1}dx$$
I tried to rationalise the denominator and got:
$$\int\frac{x(-\sqrt{1+x^2}-x^2-1)}{x^4+x^2}dx=\int\frac{-\sqrt{1+x^2}-x^2-1}{x^3+x}dx$$
But even still I find separating the integrals does not help me much in determining how to compute them, apart from:
$$\int\frac{-x^2}{x^3+x}dx$$
For the others I see no clear subsitution or trick, so any help would be great
 A: Using the substitution $ x= \tan t $ in the original integral yields

$$ \int \frac{\tan t \sec t}{1-\sec t}dt .$$

Can you finish it now?
Note: we used the identity

$$\sec^2 t= 1 + \tan^2 t. $$

A: One hint:
Use Euler substitution. Set $\sqrt{x^2+1}=t-x$ and so the main fraction will changed to $$\frac{-2}{t-1}-\frac{2}{t^2+1}$$
A: $$\int\frac{x}{\sqrt{1+x^2}-x^2-1}dx = \frac x{\sqrt{x^2 + 1} - (x^2 + 1)} \,dx $$
Rationalizing the denominator gives us $$\int \frac{x(\sqrt{x^2 + 1}+ (x^2 + 1)}{(x^2 + 1) - (x^2 + 1)^2}\,dx = \int \frac{x(\sqrt{x^2 + 1}+ x^2 + 1)}{-x^2 - x^4}\,dx$$
$$= \int \frac{\sqrt{x^2 + 1} + x^2 + 1}{-x(x^2 + 1)}\,dx = \int \frac{-1}{x\sqrt{x^2 + 1}} \,dx - \int \frac 1x\,dx$$
A: $$\int\frac{x}{\sqrt{1+x^2}-x^2-1}dx=\int\frac{x}{\sqrt{1+x^2}-(1+x^2)}dx$$
Let $u=1+x^2$ and $du=2xdx$. 
$$
\frac{1}{2}\int\frac{du}{\sqrt{u}-u}=\frac{1}{2}\int\frac{du}{(1-\sqrt{u})\sqrt{u}}.
$$
This is now the integral of a function times its derivative which means we want to use the natural logarithm.
