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integral of (x^(1/3))/(x^(1/3)+1)
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\[\int\limits \frac{\sqrt[3]{x}}{\sqrt[3]{x}+1} \, dx\]
u-sub \[u=\sqrt[3]{x} \\ \\ du=\frac{1}{3x^{2/3}}dx\] \[3 \int\limits \frac{u^3}{u+1} \, du\] Do long division then integrate separately
how did you go from a power of 1/3 in the numerator to a power of three, and what happened to the 1/(3x^(2/3))?
We have \[u=\sqrt[3]{x}...(1) \\ \\ du=\frac{1}{3x^{2/3}}dx\] \[ du=\frac{1}{3(\sqrt[3]{x})^2}dx\] From (1), \[ du=\frac{1}{3u^2}dx \\ \\ 3u^2du=dx\] ---------------------------------------- \[\int\limits\limits \frac{\sqrt[3]{x}}{\sqrt[3]{x}+1} \, dx \\ \\ \int\limits\limits \frac{u}{u+1}(3u^2) \, dx \\ \\ 3 \int\limits\limits \frac{u^3}{u+1} \, du\]
Okay, thanks a lot.
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