# What is the antiderivative of ln(root3(x))?

Jan 8, 2016

$I = \frac{1}{3} \left(\ln \left(x\right) x - x\right) + c$

#### Explanation:

Use the log property that we can pass exponents to the front of the log
$\ln \left(\sqrt[3]{x}\right) = \ln \frac{x}{3}$

$I = \int \ln \frac{x}{3} \mathrm{dx}$
$I = \frac{1}{3} \int \ln \left(x\right) \mathrm{dx}$

Say $u = \ln \left(x\right)$ so $\mathrm{du} = \frac{1}{x}$ and $\mathrm{dv} = 1$ so $v = x$

$I = \frac{1}{3} \left(\ln \left(x\right) x - \int \frac{x}{x} \mathrm{dx}\right)$
$I = \frac{1}{3} \left(\ln \left(x\right) x - \int \mathrm{dx}\right)$
$I = \frac{1}{3} \left(\ln \left(x\right) x - x\right) + c$

In general, we can say that $\int \ln \left({x}^{n}\right) \mathrm{dx} = n \left(\ln \left(x\right) - x\right) + c$