How can I solve this integral involving cube roots?

[Yankel]

Hello

I am working on this integral

\[\int \frac{\sqrt[3]{x}}{(\sqrt[3]{x}+1)^{5}}\]I have tried using a substitution, I did:

\[u=\sqrt[3]{x}+1\]

and I got that the integral becomes:

\[3\cdot \int \frac{(u-1)(u-1^{2})}{u^{5}}du\]I moved on from there, got a result, however it was not identical to the one wolfrm alpha or maple got...I am stuck...

 

[Ackbach]

Well, if $u=\sqrt[3]{x}+1$, then $u-1=\sqrt[3]{x}$, and $du=(1/3)x^{-2/3}\,dx$, or $3(x^{2/3}) \, du= dx$, and hence $3(u-1)^{2} \, du=dx$. The integral becomes
$$\int \frac{\sqrt[3]{x}}{(\sqrt[3]{x}+1)^{5}} \, dx= \int \frac{3(u-1)^{3}}{u^{5}} \, du.$$
Perhaps you have the square in the wrong place?

 

[Prove It]

Hello

I am working on this integral

\[\int \frac{\sqrt[3]{x}}{(\sqrt[3]{x}+1)^{5}}\]I have tried using a substitution, I did:

\[u=\sqrt[3]{x}+1\]

and I got that the integral becomes:

\[3\cdot \int \frac{(u-1)(u-1^{2})}{u^{5}}du\]I moved on from there, got a result, however it was not identical to the one wolfrm alpha or maple got...I am stuck...

$$\displaystyle \begin{align*} \int{ \frac{\sqrt[3]{x}}{\left( \sqrt[3]{x} + 1 \right) ^5} \, dx} &= \int{\frac{3\left( \sqrt[3]{x} \right) ^2 \, \sqrt[3]{x} }{3 \left( \sqrt[3]{x} \right) ^2 \left( \sqrt[3]{x} + 1 \right) ^5 } \, dx} \\ &= 3\int{ \left[ \frac{x}{ \left( \sqrt[3]{x} + 1 \right) ^5 } \right] \, \left[ \frac{1}{3 \left( \sqrt[3]{x} \right) ^2 } \right] \, dx} \end{align*}$$

Now let $$\displaystyle \begin{align*} u = \sqrt[3]{x} + 1 = x^{\frac{1}{3}} + 1 \implies \frac{du}{dx} = \frac{1}{3}x^{-\frac{2}{3}} \implies du = \frac{1}{3\left( \sqrt[3]{x} \right) ^2 } \, dx \end{align*}$$ and the integral becomes

$$\displaystyle \begin{align*} 3\int{ \left[ \frac{x}{ \left( \sqrt[3]{x} + 1 \right) ^5} \right] \, \left[ \frac{1}{3 \left( \sqrt[3]{x} \right) ^2} \right] \, dx} &= 3\int{ \frac{(u - 1)^3}{u^5} \, du } \\ &= 3 \int{ \frac{u^3 - 3u^2 + 3u - 1}{u^5} \, du} \\ &= 3\int{ u^{-2 }- 3u^{-3} + 3u^{-4} - u^{-5} \, du} \end{align*}$$

You should be able to integrate this now :)