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 ?
 A: 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 $dv = \dfrac x{\sqrt{1 + x^2}}\implies v = \sqrt{1 + x^2}$
$$I = uv - \int u'v$$
A: $\int\dfrac{x\sin^{-1}x}{\sqrt{1+x^2}}dx$
$=\int\dfrac{\sin^{-1}x}{2\sqrt{1+x^2}}d(x^2)$
$=\int\sin^{-1}x~d(\sqrt{1+x^2})$
$=\sqrt{1+x^2}\sin^{-1}x-\int\sqrt{1+x^2}~d(\sin^{-1}x)$
$=\sqrt{1+x^2}\sin^{-1}x-\int\dfrac{\sqrt{1+x^2}}{\sqrt{1-x^2}}dx$
$=\sqrt{1+x^2}\sin^{-1}x-\int\dfrac{\sqrt{1+\sin^2u}}{\sqrt{1-\sin^2u}}d(\sin u)$ $(\text{Let}~x=\sin u)$
$=\sqrt{1+x^2}\sin^{-1}x-\int\sqrt{1+\sin^2u}~du$
$=\sqrt{1+x^2}\sin^{-1}x-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{2n}u}{4^n(n!)^2(1-2n)}du$
$=\sqrt{1+x^2}\sin^{-1}x-\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!\sin^{2n}u}{4^n(n!)^2(1-2n)}\right)du$
For $\int\sin^{2n}u~du$ , where $n$ is any natural number,
$\int\sin^{2n}u~du=\dfrac{(2n)!u}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts from the formula of http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions.
$\therefore\sqrt{1+x^2}\sin^{-1}x-\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!\sin^{2n}u}{4^n(n!)^2(1-2n)}\right)du$
$=\sqrt{1+x^2}\sin^{-1}x-u-\sum\limits_{n=1}^\infty\dfrac{(-1)^n((2n)!)^2u}{16^n(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2\sin^{2k-1}u\cos u}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)}+C$
$=\sqrt{1+x^2}\sin^{-1}x-\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2u}{16^n(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2\sin^{2k-1}u\cos u}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)}+C$
$=\sqrt{1+x^2}\sin^{-1}x-\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2\sin^{-1}x}{16^n(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2x^{2k-1}\sqrt{1-x^2}}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)}+C$
A: 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 $$I=\int\phi\sin\phi d\phi=\phi\int\sin\phi d\phi-\int\left(\frac{d\phi}{d\phi}\cdot\int\sin\phi d\phi\right)d\phi $$
$$=-\phi\cdot\cos\phi+\int \cos\phi d\phi=\cdots$$
