# How do I evaluate int (arcsin x)/sqrt (1+x) dx?

Feb 17, 2015

We will integrate this using integration by parts.

Recall that $u v - \int v \mathrm{du}$

Let $u = \arcsin \left(x\right)$ and $\mathrm{dv} = \frac{\mathrm{dx}}{\sqrt{1 + x}}$

Differentiating $u = \arcsin x$ we have
$\mathrm{du} = \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Integrate $\mathrm{dv} = \frac{\mathrm{dx}}{\sqrt{1 + x}}$

$\int \mathrm{dv} = \int \frac{\mathrm{dx}}{\sqrt{1 + x}}$

$v = 2 \sqrt{1 + x}$

Therefore,

$\arcsin \left(x\right) \left(2 \sqrt{1 + x}\right) - \int 2 \sqrt{1 + x} \left(\frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

Some rewriting and factor $1 - {x}^{2}$

$2 \arcsin \left(x\right) \sqrt{1 + x} - 2 \int \frac{\sqrt{1 + x}}{\sqrt{\left(1 + x\right) \left(1 - x\right)}} \mathrm{dx}$

$2 \arcsin \left(x\right) \sqrt{1 + x} - 2 \int \frac{\sqrt{1 + x}}{\sqrt{1 + x} \sqrt{1 - x}} \mathrm{dx}$

$2 \arcsin \left(x\right) \sqrt{1 + x} - 2 \int \frac{1}{\sqrt{1 - x}} \mathrm{dx}$

Finally, we integrate

$2 \arcsin \left(x\right) \sqrt{1 + x} + 4 \sqrt{1 - x} + C$