Integrate $\int_0^1 \sin^{-1}{\frac{x^2}{1+x^2}}dx$ 
Integrate $\int_0^1 \sin^{-1}{\frac{x^2}{1+x^2}}dx$

I tried to put $x=\tan\theta$, which gives
$\int_0^{\frac{\pi}{4}} {\sin^{-1}({\sin^2\theta}})\sec^2\theta d\theta$, but I don't know how to proceed after this.
Is there something I am missing here?
 A: Hint: 
You might know the property that $$\int_a^b f(x) dx+\int_{f(a)}^{f(b)} f^{-1} (x) dx=bf(b)-af(a)$$
Using this we get $$\int_0^1 \arcsin \left( \frac {x^2}{1+x^2}\right) dx+\int_0^{\frac {\pi}{6}} \sqrt {\frac {\sin x}{1-\sin x}} dx=\frac {\pi}{6}$$
Therefore $$\int_0^1 \arcsin \left( \frac {x^2}{1+x^2}\right) dx=\frac {\pi}{6}- \int_0^{\frac {\pi}{6}} \sqrt {\frac {\sin x}{1-\sin x}} dx$$ 
Now for the integral on right hand side use the  substitution $u=\sin x$ which will lead you to a very simple integral $$\int_0^{\frac 12} \sqrt {\frac {u}{1-u}} \frac {du}{\sqrt {1-u^2}}$$ Which can be solved as stated by Jack D'Aurizio
A: $$\int_{0}^{1}\arcsin\left(\frac{x^2}{1+x^2}\right)\,dx = \int_{0}^{1}\arcsin\left(\frac{x}{1+x}\right)\frac{dx}{2\sqrt{x}} $$
equals
$$ \int_{0}^{1/2}\frac{\arcsin(u)}{2u^{1/2}(1-u)^{3/2}}\,du $$
which (pretty incredibly) can be managed by integration by parts. It boils down to
$$ \frac{\pi}{6}-\int_{0}^{1/2}\sqrt{\frac{u}{1-u}}\cdot\frac{du}{\sqrt{1-u^2}}\,du $$
then to
$$ \color{red}{\frac{\pi}{6}+\log(2+\sqrt{3})-\sqrt{2}\log(\sqrt{2}+\sqrt{3})}\approx 0.219563.$$
