<s>[INST] Find the integral $\int \frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx$ $$\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}=\frac{2\sqrt[5]{2x-3}-1}{2\sqrt[5]{2x-3}(x-1)}=\frac{1}{x-1}-\frac{1}{2\sqrt[5]{2x-3}(x-1)}$$
$$\int \frac{1}{x-1}\mathrm dx=\ln|x-1|+c$$
$$\int \frac{1}{2\sqrt[5]{2x-3}(x-1)}\mathrm dx=\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx$$
Substitution $u=2x-3,du=2dx$ gives
$$\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du$$
How to evaluate $\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du$?
 [/INST] As @Archis Welankar noted, this integral is familiar. If we let $u=(2x-3)^{\frac15}$, then $2x-3=u^5$, so $dx=\frac52du$. Then
$$\int\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}dx=\int\frac{2u-1}{u^6+u}\cdot\frac52u^4du=\frac52\int\frac{2u^4-u^3}{u^5+1}du$$
I calculated this integral a couple of weeks ago. Today I checked that integral by comparing with numerical solution, and it passed verification.
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