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.
 A: Use $x=t^6$ as a new substitution to integrate the first integral. So you have $$\int\frac{6t^5 dt}{t^2(t+5)}=\int\frac{6t^3dt}{t+5}=\int6\left( t^2-5t+25-\frac{125}{t+5}\right)dt$$ which is elementary. Whenever you have an integral like $$\int R(x,x^{\frac{p_1}{q_1}},...,x^{\frac{p_k}{q_k}})dx$$ you can use $x=t^{\text{lcm}({q_1},...{q_k})}$ as a substitution.
A: $$
\int \frac{1}{u(u-5)^3}du=\int\frac{A}{u}+\frac{B}{(u-5)}+\frac{C}{(u-5)^2}+\frac{D}{(u-5)^3}du.
$$
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:
$$
\frac{7}{6u(u-5)^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{6}})^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{2}})}=\frac{7}{6}\frac{1}{(5x^{\frac{1}{2}}+x^{\frac{2}{3}})}\ne \frac{1}{x^{\frac{1}{2}}+5x^{\frac{1}{3}}}.
$$
A: Putting $x=(t-5)^6, dx=6(t-5)^5$ and $x=0\implies t=5,x=1\implies t=6$
We get $$ \int_0^1 \frac{dx}{x^\frac12+5x^\frac13}=\int_5^6 \frac{6(t-5)^3dt}{t} =6\int_5^6 (t^2-15t+75-125\frac1t)dt=6(\frac{t^3}3-15\frac{t^2}2+75t-125\log t)\mid_5^6$$
$$=2(6^3-5^3)-45(6^2-5^2)+450(6-5)-750\log\frac65$$
