# How do you evaluate the integral int dx/(root3(x)+1)?

Apr 23, 2018

The integral is equal to $\frac{3}{2} \left({x}^{\frac{2}{3}} - 2 {x}^{\frac{1}{3}} + 2 \ln | {x}^{\frac{1}{3}} + 1 |\right) + C$.

#### Explanation:

$\textcolor{w h i t e}{=} \int \frac{1}{\sqrt[3]{x} + 1}$ $\mathrm{dx}$

$= \int \frac{1}{{x}^{\frac{1}{3}} + 1}$ $\mathrm{dx}$

To solve the integral, substitute $u = {x}^{\frac{1}{3}} + 1$, which means:

$\mathrm{du} = \frac{1}{3} {x}^{- \frac{2}{3}} \mathrm{dx}$

$\mathrm{du} = \frac{1}{3 {x}^{\frac{2}{3}}} \mathrm{dx}$

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

But we can solve for ${x}^{\frac{2}{3}}$ using our original substitution:

$u = {x}^{\frac{1}{3}} + 1$

$u - 1 = {x}^{\frac{1}{3}}$

${\left(u - 1\right)}^{2} = {x}^{\frac{2}{3}}$

Put this in the other equation:

$3 {\left(u - 1\right)}^{2} \mathrm{du} = \mathrm{dx}$

Plug this into the integral:

$\textcolor{w h i t e}{=} \int \frac{1}{u} \cdot 3 {\left(u - 1\right)}^{2} \mathrm{du}$

$= 3 \int {\left(u - 1\right)}^{2} / u$ $\mathrm{du}$

$= 3 \int \frac{{u}^{2} - 2 u + 1}{u}$ $\mathrm{du}$

$= 3 \int \left(u - 2 + \frac{1}{u}\right)$ $\mathrm{du}$

=3(intu $\mathrm{du} - \int 2$ du+int1/udu)

$= 3 \left({u}^{2} / 2 - 2 u + \ln | u |\right) + C$

$= \frac{3 {u}^{2}}{2} - 6 u + 3 \ln | u | + C$

$= \frac{3 {\left({x}^{\frac{1}{3}} + 1\right)}^{2}}{2} - 6 \left({x}^{\frac{1}{3}} + 1\right) + 3 \ln | {x}^{\frac{1}{3}} + 1 | + C$

$= \frac{3 \left({x}^{\frac{2}{3}} + 2 {x}^{\frac{1}{3}} + 1\right)}{2} - 6 \left({x}^{\frac{1}{3}} + 1\right) + 3 \ln | {x}^{\frac{1}{3}} + 1 | + C$

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

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

$= \frac{3 {x}^{\frac{2}{3}} + 6 {x}^{\frac{1}{3}} - 12 {x}^{\frac{1}{3}} + 6 \ln | {x}^{\frac{1}{3}} + 1 |}{2} + C$

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

$= \frac{3 \left({x}^{\frac{2}{3}} - 2 {x}^{\frac{1}{3}} + 2 \ln | {x}^{\frac{1}{3}} + 1 |\right)}{2} + C$

$= \frac{3}{2} \left({x}^{\frac{2}{3}} - 2 {x}^{\frac{1}{3}} + 2 \ln | {x}^{\frac{1}{3}} + 1 |\right) + C$

That's the integral. Hope this helped!