integrate $\int\frac{\sin x}{1+\sin^{2}x}dx$ 
$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}$$

$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}=\int\frac{\sin x}{2-\cos^{2}x}\mathrm {dx}$$
$u=\cos x$
$du=-\sin x dx$
$$-\int\frac{\mathrm {du}}{2-u^{2}}$$
How could I continue?

Hint:
Write $$\frac{-1}{2-u^2}=\frac{-1}{(\sqrt{2}+u)(\sqrt{2}-u)}=\frac{-1}{2\sqrt{2}}\left(\frac{1}{\sqrt{2}+u}+\frac{1}{\sqrt{2}-u}\right)$$
