# How do you find the integral of int (4x+3)/sqrt(1-x^2)?

Jan 3, 2017

The answer is $= - 4 \sqrt{1 - {x}^{2}} + 3 \arcsin x + C$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

${\sin}^{2} x + {\cos}^{2} x = 1$

Rewrite the integral (we apply linearity)

$\int \frac{\left(4 x + 3\right) \mathrm{dx}}{\sqrt{1 - {x}^{2}}} = \int \frac{4 x \mathrm{dx}}{\sqrt{1 - {x}^{2}}} + \int \frac{3 \mathrm{dx}}{\sqrt{1 - {x}^{2}}}$

$= 4 \int \frac{x \mathrm{dx}}{\sqrt{1 - {x}^{2}}} + 3 \int \frac{\mathrm{dx}}{\sqrt{1 - {x}^{2}}}$

Let $u = 1 - {x}^{2}$, $\implies$, $\mathrm{du} = - 2 x \mathrm{dx}$

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

$= - \frac{1}{2} \frac{\sqrt{u}}{\frac{1}{2}} = - \sqrt{u} = - \sqrt{1 - {x}^{2}}$

Let $x = \sin \theta$, $\implies$, $\mathrm{dx} = \cos \theta \left(d \theta\right)$

$\sqrt{1 - {x}^{2}} = \sqrt{1 - {\sin}^{2} \theta} = \sqrt{{\cos}^{2} \theta} = \cos \theta$

$\int \frac{\mathrm{dx}}{\sqrt{1 - {x}^{2}}} = \int \frac{\cos \theta d \theta}{\cos} \theta = \int d \theta = \theta$

$= \arcsin x$

Putting it all together

$\int \frac{\left(4 x + 3\right) \mathrm{dx}}{\sqrt{1 - {x}^{2}}} = - 4 \sqrt{1 - {x}^{2}} + 3 \arcsin x + C$