# How do you find the indefinite integral of int root3x/(root3x-1)?

Mar 20, 2018

${\left(\sqrt[3]{x} - 1\right)}^{3} + \frac{9 {\left(\sqrt[3]{x} - 1\right)}^{2}}{2} + 9 \left(\sqrt[3]{x} - 1\right) + 3 \ln \left(\left\mid \sqrt[3]{x} - 1 \right\mid\right) + C$

#### Explanation:

We have $\int \frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1} \mathrm{dx}$

Substitute $u = \left(\sqrt[3]{x} - 1\right)$
$\frac{\mathrm{du}}{\mathrm{dx}} = {x}^{- \frac{2}{3}} / 3$

$\mathrm{dx} = 3 {x}^{\frac{2}{3}} \mathrm{du}$

$\int \frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1} \left(3 {x}^{\frac{2}{3}}\right) \mathrm{du} = \int \frac{3 x}{\sqrt[3]{x} - 1} \mathrm{du} = \int \frac{3 {\left(u + 1\right)}^{3}}{u} \mathrm{du} = 3 \int \frac{{u}^{3} + 3 {u}^{2} + 3 u + 1}{u} \mathrm{du} = \int 3 {u}^{2} + 9 u + 9 + \frac{3}{u} \mathrm{du} = {u}^{3} + \frac{9 {u}^{2}}{2} + 9 u + 3 \ln \left(\left\mid u \right\mid\right) + C$

Resubstitute $u = \sqrt[3]{x} - 1$:
${\left(\sqrt[3]{x} - 1\right)}^{3} + \frac{9 {\left(\sqrt[3]{x} - 1\right)}^{2}}{2} + 9 \left(\sqrt[3]{x} - 1\right) + 3 \ln \left(\left\mid \sqrt[3]{x} - 1 \right\mid\right) + C$