# What is the integration of (dx)/(x.sqrt(x^3+4))??

Mar 7, 2018

$\frac{1}{6} \ln | \frac{\sqrt{{x}^{3} + 4} - 2}{\sqrt{{x}^{3} + 4} + 2} | + C$

#### Explanation:

Substitute ${x}^{3} + 4 = {u}^{2}$. Then $3 {x}^{2} \mathrm{dx} = 2 u \mathrm{du}$, so that

$\frac{\mathrm{dx}}{x \sqrt{{x}^{3} + 4}} = \frac{2 u \mathrm{du}}{3 {x}^{3} u} = \frac{2}{3} \frac{\mathrm{du}}{{u}^{2} - 4} = \frac{1}{6} \left(\frac{\mathrm{du}}{u - 2} - \frac{\mathrm{du}}{u + 2}\right)$

Thus

$\int \frac{\mathrm{dx}}{x \sqrt{{x}^{3} + 4}} = \frac{1}{6} \int \left(\frac{\mathrm{du}}{u - 2} - \frac{\mathrm{du}}{u + 2}\right) = \frac{1}{6} \ln | \frac{u - 2}{u + 2} | + C$
$= \frac{1}{6} \ln | \frac{\sqrt{{x}^{3} + 4} - 2}{\sqrt{{x}^{3} + 4} + 2} | + C$