# How do you integrate (x^(1/3))/(((x^(1/3))-1))?

Jan 7, 2017

$x + \frac{3}{2} {x}^{\frac{2}{3}} + 3 {x}^{\frac{1}{3}} + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid + C$

#### Explanation:

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

Let $u = {x}^{\frac{1}{3}} - 1$. This implies that ${x}^{\frac{1}{3}} = u + 1$ and that $x = {\left(u + 1\right)}^{3}$, a fact we use to show that $\mathrm{dx} = 3 {\left(u + 1\right)}^{2} \mathrm{du}$.

Substituting in what we know, this becomes:

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

Expanding, then dividing:

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

Integrating term by term:

$I = 3 \left({u}^{3} / 3 + \frac{3}{2} {u}^{2} + 3 u + \ln \left\mid u \right\mid\right)$

$I = {u}^{3} + \frac{9}{2} {u}^{2} + 9 u + 3 \ln \left\mid u \right\mid$

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

$I = {\left({x}^{\frac{1}{3}} - 1\right)}^{3} + \frac{9}{2} {\left({x}^{\frac{1}{3}} - 1\right)}^{2} + 9 \left({x}^{\frac{1}{3}} - 1\right) + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid$

If you wish, you can expand all of these terms and combine for the simplified answer of:

$I = \left(x - 3 {x}^{\frac{2}{3}} + 3 {x}^{\frac{1}{3}} - 1\right) + \frac{9}{2} \left({x}^{\frac{2}{3}} - 2 {x}^{\frac{1}{3}} + 1\right) + 9 {x}^{\frac{1}{3}} - 9 + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid$

Continue with combining like terms, and have the $- 9 , 1 ,$ and $\frac{9}{2}$ absorb into the constant of integration:

$I = x + \frac{3}{2} {x}^{\frac{2}{3}} + 3 {x}^{\frac{1}{3}} + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid + C$

Jan 8, 2017

I also got $x + 3 {x}^{\text{1/3" + 3/2x^"2/3" + 3ln|x^"1/3}} - 1 | + C$.

Another way to do it is:

int x^"1/3"/(x^"1/3" - 1)dx

= int cancel((x^"1/3" - 1)/(x^"1/3" - 1))^(1) + 1/(x^"1/3" - 1)dx

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

For this, let $u = {x}^{\text{1/3}}$ so that $\mathrm{du} = \frac{1}{3} {x}^{- \text{2/3}} \mathrm{dx}$, or $\mathrm{dx} = 3 {x}^{\text{2/3}} \mathrm{du} = 3 {u}^{2} \mathrm{du}$. Then:

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

Dividing the second integrand gives:

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

so, this overall gives:

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

$= {u}^{3} + 3 u + \frac{3}{2} {u}^{2} + 3 \ln | u - 1 |$

$= \textcolor{b l u e}{x + 3 {x}^{\text{1/3" + 3/2x^"2/3" + 3ln|x^"1/3}} - 1 | + C}$

Jan 8, 2017

Making $y = {x}^{\frac{1}{3}}$ we have $\mathrm{dy} = \frac{1}{3} {x}^{- \frac{2}{3}} \mathrm{dx} = \frac{1}{3} \frac{\mathrm{dx}}{y} ^ 2$ so

${x}^{\frac{1}{3}} / \left({x}^{\frac{1}{3}} - 1\right) \mathrm{dx} = 3 {y}^{3} / \left(y - 1\right) \mathrm{dy} = 3 \left(\frac{{y}^{3} - 1}{y - 1} + \frac{1}{y - 1}\right) \mathrm{dy} = 3 \left(1 + y + {y}^{2} + \frac{1}{y - 1}\right) \mathrm{dy}$

so

$\int {x}^{\frac{1}{3}} / \left({x}^{\frac{1}{3}} - 1\right) \mathrm{dx} = 3 \int \left(1 + y + {y}^{2} + \frac{1}{y - 1}\right) \mathrm{dy} = 3 y + \frac{3}{2} {y}^{2} + {y}^{3} + \log \left(\left\mid y - 1 \right\mid\right) + C$ or

$\int {x}^{\frac{1}{3}} / \left({x}^{\frac{1}{3}} - 1\right) \mathrm{dx} = 3 {x}^{\frac{1}{3}} + \frac{3}{2} {x}^{\frac{2}{3}} + x + 3 \log \left(\left\mid {x}^{\frac{1}{3}} - 1 \right\mid\right) + C$