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 ?

HINT:
Let $x^2=\cos2y$ as $x^2\ge0,0\le2y\le\dfrac\pi2\  \ \ \ (1)$
$\implies x\ dx=-\sin2y\ dy,\sin2y=+\sqrt{1-x^4}$
Now $\sin2y=2\sin y\cos y$
and $\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y\text{ by } (1)$
Can you take it from here? 
