# Integrate?

## $\frac{2 x + 1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Apr 11, 2018

$\int \frac{2 x + 1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$ = $\int \frac{2 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} + \int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

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

#### Explanation:

You need to separate the integral first, then do a u-substitution on the side that has a 2x in the numerator and you need to use the arcsin integral trig formula to get the side with a 1 in the numerator.

Apr 11, 2018

#### Explanation:

$I = \int$ $\frac{2 x + 1}{\sqrt{1 - {x}^{2}}}$ $\mathrm{dx}$

$= \int$ $\frac{2 x}{\sqrt{1 - {x}^{2}}} + \frac{1}{\sqrt{1 - {x}^{2}}}$ $\mathrm{dx}$

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

$J = \int$ $\frac{2 x}{\sqrt{1 - {x}^{2}}}$ $\mathrm{dx}$

Let:

$\sin \left(u\right) = x$
$\cos \left(u\right)$ $\mathrm{du}$ = $\mathrm{dx}$

$J = \int$ $\frac{2 \sin \left(u\right) \cos \left(u\right)}{\sqrt{1 - {\sin}^{2} \left(u\right)}}$ $\mathrm{du}$

$= \int$ $\frac{2 \sin \left(u\right) \cos \left(u\right)}{\sqrt{{\cos}^{2} \left(u\right)}}$ $\mathrm{du}$

$= \int$ $2 \sin \left(u\right)$ $\mathrm{du}$

$= - 2 \cos \left(u\right) + C$

$= - 2 \cos \left(\arcsin \left(x\right)\right) + C$

Therefore:

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

This can be simplified further:

Let $\theta = \arcsin \left(x\right)$:
$\sin \left(\theta\right) = x$
$\cos \left(\theta\right) = \sqrt{1 - {x}^{2}}$

$\cos \left(\arcsin \left(x\right)\right) = \sqrt{1 - {x}^{2}}$

Finally:

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