Evaluating $\int\frac{x\sin^{-1}(x)}{\sqrt{1+x^{2}}}\mathrm{d}x$ How can we evaluate:

$$\int\frac{x\sin^{-1}(x)}{\sqrt{1+x^{2}}}\mathrm{d}x$$

I tried to use integration by parts, but the positive sign of the $x^2$ in the square root doesn't cancel with the integral of $\sin^{-1}(x)$, so how can I proceed ?
Yes, integration by parts is the route to go...

Try putting $u = \sin^{-1} x \implies du = \dfrac {dx}{\sqrt{1-x^2}}$

and put $dv = \dfrac x{\sqrt{1 + x^2}}\implies v = \sqrt{1 + x^2}$

$$I = uv - \int u'v$$
As the existing integral is not elementary 

I assume the Question to be $$I=\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx$$

Setting  $\sin^{-1}x=\phi\implies \sin\phi=x, \cos\phi=+\sqrt{1-x^2}$ as the principal value of $\sin^{-1}x$ lies in $\left[-\frac\pi2,\frac\pi2\right]$

$$I=\int\phi\sin\phi d\phi $$

Now integrate by parts $$I=\int\phi\sin\phi d\phi=\phi\int\sin\phi d\phi-\int\left(\frac{d\phi}{d\phi}\cdot\int\sin\phi d\phi\right)d\phi $$

$$=-\phi\cdot\cos\phi+\int \cos\phi d\phi=\cdots$$