Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$. 
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$


Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J =  2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\left(\sqrt{1+ x^2}\over 2\right) + {\sqrt{1-x^4}\over 2} + C $$.
The give answer is $\displaystyle \arcsin(x^2) + {\sqrt{1-x^4}\over 2} + C$.
What went wrong in my attempt ?

$u = \sqrt {1+x^2}$
$u^2 = 1 + x^2$
$2u du = 2x dx$
$u du = x dx$
Here you are wrong.
