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.

Putting $x=(t-5)^6, dx=6(t-5)^5$ and $x=0\implies t=5,x=1\implies t=6$
We get $$ \int_0^1 \frac{dx}{x^\frac12+5x^\frac13}=\int_5^6 \frac{6(t-5)^3dt}{t} =6\int_5^6 (t^2-15t+75-125\frac1t)dt=6(\frac{t^3}3-15\frac{t^2}2+75t-125\log t)\mid_5^6$$
$$=2(6^3-5^3)-45(6^2-5^2)+450(6-5)-750\log\frac65$$
