# How do you find the integral of (2x-3)/((9-x^2)^0.5)dx?

##### 1 Answer
Aug 5, 2015

$- 2 \sqrt{9 - {x}^{2}} - 3 \arcsin \left(\frac{x}{3}\right) + C$

#### Explanation:

Start off by rewriting as follows

$\int \frac{2 x}{9 - {x}^{2}} ^ \left(\frac{1}{2}\right) \mathrm{dx} - \int \frac{3}{9 - {x}^{2}} ^ \left(\frac{1}{2}\right) \mathrm{dx}$

For the first integral we use a u substitution

Let $u = 9 - {x}^{2}$ then $\frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$

$\mathrm{dx} = \frac{\mathrm{du}}{- 2 x}$

Make the substitution

$- \int \frac{2 x}{u} ^ \left(\frac{1}{2}\right) \frac{\mathrm{du}}{2 x} = - \int {u}^{- \frac{1}{2}} \mathrm{du}$

Integrating we have

$- 2 {u}^{\frac{1}{2}}$

Back substituting for u

$- 2 {\left(9 - {x}^{2}\right)}^{\frac{1}{2}}$

$- 2 \sqrt{9 - {x}^{2}}$

For the second integral we will use a trigonometric substitution

Let $x = 3 \sin \theta$ the

$\mathrm{dx} = 3 \cos \theta d \theta$

Make the substitution

$\int \frac{3}{9 - 9 {\sin}^{2} \theta} ^ \left(\frac{1}{2}\right) 3 \cos d \theta$

$\int \frac{9 \cos \theta}{9 \left[1 - {\sin}^{2} \theta\right]} ^ \left(\frac{1}{2}\right) d \theta$

$3 \int \cos \frac{\theta}{{\cos}^{2} \theta} ^ \left(\frac{1}{2}\right) d \theta$

$3 \int \cos \frac{\theta}{\cos} \theta d \theta$

$3 \int d \theta$

Integrating we get $\theta$

Now $\theta = \arcsin \left(\frac{x}{3}\right)$

So we have $3 \arcsin \left(\frac{x}{3}\right)$

Putting the two results together we have

$- 2 \sqrt{9 - {x}^{2}} - 3 \arcsin \left(\frac{x}{3}\right) + C$