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 ?

Put $t=x^2$
and
$$\frac {1-t}{1+t}=u^2$$
or
$$t=\frac{1-u^2}{1+u^2} $$
$$=\frac {2}{1+u^2}-1$$
and
$$dt=\frac{-4u}{(1+u^2)^2} $$
