<s>[INST] How to integrate $(x^{1/2} + 5 x^{1/3})^{-1}$? $$ \int_0^1 \frac{dx}{x^\frac12+5x^\frac13} $$
Let $$u = 5+x^\frac16 $$
I have change x into u, and find $$\int_5^6 \frac{7}{6u(u-5)^3} $$
Then, i don't know how to integrate it.
 [/INST] $$
\int \frac{1}{u(u-5)^3}du=\int\frac{A}{u}+\frac{B}{(u-5)}+\frac{C}{(u-5)^2}+\frac{D}{(u-5)^3}du.
$$
Courtesy of partial fraction decomposition. :-)

Edit: It would appear your $u$ sub does not work. (Thanks, Gerry.) Nonetheless, I hope this is a helpful illustration of partial fraction decomposition.
Here is why your $u$ sub does not work: If $u=5+x^{\frac{1}{6}}$ and we wish to write $\dfrac{7}{6u(u-5)^3}$, we find the following:
$$
\frac{7}{6u(u-5)^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{6}})^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{2}})}=\frac{7}{6}\frac{1}{(5x^{\frac{1}{2}}+x^{\frac{2}{3}})}\ne \frac{1}{x^{\frac{1}{2}}+5x^{\frac{1}{3}}}.
$$
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