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I've tried it several times and can't seem to get it, always get stuck midway through

$$\int\frac{(x)*arcsin(x)}{\sqrt {1-x^2}} dx $$

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$u = x, du = dx\\ dv = \frac {\sin^{-1} x}{\sqrt{1-x^2}} dx$

$v = \int \frac {\sin^{-1} x}{\sqrt{1-x^2}}dx\\ s = \sin^{-1} x, ds = \frac 1{\sqrt{1-x^2}} dx$

$v = \frac 12 (\sin^{-1} x)^2$

$x(\sin^{-1}x)^2 - \int (\sin^{-1}x)^2 \;dx$

that looks like a dead end... how about:

$u = \sin^{-1} x, du = \frac 1{\sqrt {1-x^2}} \;dx\\ dv = \frac {x}{\sqrt{1-x^2}} \;dx, v = -\sqrt{1-x^2}$

$-(\sin^{-1} x)\sqrt{1-x^2} + \int dx$

That looks promissing

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  • $\begingroup$ Yeah thanks :) I was doing it just like that and didn't get anywhere... $\endgroup$ – TatiiLourenco Oct 31 '16 at 18:54
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Using integration by parts:

$$\int f(x)g'(x)\space\text{d}x=f(x)g(x)-\int f'(x)g(x)\space\text{d}x$$

So:

$$\mathcal{I}\left(x\right)=\int\frac{x\arcsin\left(x\right)}{\sqrt{1-x^2}}\space\text{d}x=-\arcsin(x)\sqrt{1-x^2}+\int1\space\text{d}x$$

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