# How do I evaluate intsqrt(4x^2-9)/xdx?

Jun 22, 2018

The answer is $= \sqrt{4 {x}^{2} - 9} - 3 \arctan \left(\frac{1}{3} \sqrt{4 {x}^{2} - 9}\right) + C$

#### Explanation:

Let $u = \sqrt{4 {x}^{2} - 9}$, then

$\mathrm{du} = \frac{8 x \mathrm{dx}}{2 \sqrt{4 {x}^{2} - 9}} = \frac{4 x \mathrm{dx}}{\sqrt{4 {x}^{2} - 9}}$

The integral is

$I = \int \frac{\sqrt{4 {x}^{2} - 9} \mathrm{dx}}{x}$

$= \int \frac{\mathrm{du} \sqrt{4 {x}^{2} - 9} \cdot \sqrt{4 {x}^{2} - x}}{4 x \cdot x}$

$= \int \frac{{u}^{2} \mathrm{du}}{{u}^{2} + 9}$

$= \int \frac{\left({u}^{2} + 9\right) \mathrm{du}}{{u}^{2} + 9} - \int \frac{9 \mathrm{du}}{{u}^{2} + 9}$

$= \int \mathrm{du} - 9 \int \frac{\mathrm{du}}{{u}^{2} + 9}$

$= u - 9 \int \frac{\mathrm{du}}{{u}^{2} + 9}$

${I}_{1} - {I}_{2}$

The second integral is

${I}_{2} = 9 \int \frac{\mathrm{du}}{{u}^{2} + 9}$

Let $v = \frac{u}{3}$, $\implies$, $\mathrm{dv} = \frac{\mathrm{du}}{3}$

${I}_{2} = 9 \int \frac{3 \mathrm{dv}}{9 {v}^{2} + 9}$

$= \frac{27}{9} \int \frac{\mathrm{dv}}{{v}^{2} + 1}$

$= 3 \arctan \left(v\right)$

$= 3 \arctan \left(\frac{u}{3}\right)$

Finally the integral is

$I = u - 3 \arctan \left(\frac{u}{3}\right)$

$= \sqrt{4 {x}^{2} - 9} - 3 \arctan \left(\frac{1}{3} \sqrt{4 {x}^{2} - 9}\right) + C$