How to solve this irrational integral I want to solve the integral
$$\large\int\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx$$
I've gone with the varialbe change method using
 $$ 
2x-3=u\
$$
and got 
$$
\frac{1}{2}\int \frac{2u^\frac{1}{5}-1}{uu^\frac{1}{5}+u^\frac{1}{5}} du\
$$
then I used shift
$$
u=t^5
$$
and got
$$
\frac{5}{2}\int\frac{2t^4-t^3}{t^5+1} dt\
$$
This is the part that I'm stuck on.
Any kind of advise would be helpfull.
 A: HINT:
$$\int\frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\space\text{d}x=2\int\frac{\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\space\text{d}x=$$

Substitute $u=2x-3$ and $\text{d}u=2\space\text{d}x$:

$$\int\frac{\sqrt[5]{u}-1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\int\left[\frac{\sqrt[5]{u}}{u\sqrt[5]{u}+\sqrt[5]{u}}-\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\right]\space\text{d}u=$$
$$\int\frac{\sqrt[5]{u}}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\int\frac{1}{1+u}\space\text{d}u-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$

Substitute $s=1+u$ and $\text{d}s=\text{d}u$:

$$\int\frac{1}{s}\space\text{d}s-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\ln\left|s\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$
$$\ln\left|1+u\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\ln\left|1+2x-3\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$
$$\ln\left|2x-2\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$

Substitute $p=\sqrt[5]{u}$ and $\text{d}p=\frac{1}{5\sqrt[5]{u^4}}\space\text{d}u$:

$$\ln\left|2x-2\right|-5\int\frac{p^4}{p^6+p}\space\text{d}p=\ln\left|2x-2\right|-5\int\frac{p^3}{p^5+1}\space\text{d}p$$
