# What is the integral of (arcsin x)/sqrt (1+x) dx?

Feb 23, 2015

Hello,

$\int \arcsin \frac{x}{\sqrt{1 + x}} \mathrm{dx} = 2 \arcsin \left(x\right) \sqrt{1 + x} + 4 \sqrt{1 - x} + c$,
where $c \in \mathbb{R}$.

Use integration by parts formula :

$\int u ' v = u v - \int u v '$

You can prove that if you write $\left(u v\right) ' = u ' v + u v '$, and integrate that.

Here, $u \left(x\right) = \arcsin \left(x\right)$ and $v \left(x\right) = 2 \sqrt{1 + x}$. You have :
$u ' \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$ and $v ' \left(x\right) = \frac{1}{\sqrt{1 + x}}$.
Apply the formula :

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

Simplify $\frac{\sqrt{1 + x}}{\sqrt{1 - {x}^{2}}} = \frac{\sqrt{1 + x}}{\sqrt{\left(1 - x\right) \left(1 + x\right)}} = \frac{1}{\sqrt{1 - x}}$.
So, you can write

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

Finally, you know that $\int \frac{1}{\sqrt{1 - x}} \mathrm{dx} = \int {\left(1 - x\right)}^{- \frac{1}{2}} \mathrm{dx} = - 2 {\left(1 - x\right)}^{\frac{1}{2}} + c$ and the result is proved !