Problem with indefinite integration I introduce shift $t^5=2x -3$,but it doesn't help $$\int \frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}dx$$
 A: $$J:=\int \frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}dx=\int\frac{dx}{x-1}-\frac12\int\frac{dx}{(x-1)\sqrt[5]{2x-3}}=$$
Substitute in the second integral as you did:
$$t^5=2x-3\implies dx=\frac{5t^4}2dt\implies J=\log|x-1|-\frac54\int\frac{t^4\,dt}{\frac{t^5+1}2t}=$$
$$=\log|x-1|-\frac52\int\frac{t^3}{t^5+1}dt$$
Now do partial fractions:
$$t^5+1=(t+1)(t^4-t^3+t^2-t+1)$$
To decompose that you better know the roots of unity of degree $\;5\;$
A: Starting from
$$I=\int \frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\text dx$$
One substitution approach is $u=\sqrt[5]{(2x-3)^4},\,\text du=\frac 8{5\sqrt[5]{2x-3}}\text dx,$ which yields
$$I=\frac 58\int\frac{2u^\frac 14-1}{u^\frac 54+1}\text du$$
If $v^4 = u,4v^3\text dv=\text du$, then we get
$$I=\frac 52\int\frac{4v^3(2v-1)}{v^5+1}\text dv$$
With some partial fraction decomposition, we get
$$\frac 1{v^5+1}=\frac 1{v(v+1)}\left(\frac 1{v^2-\varphi v+1}+\frac 1{v^2+\frac v\varphi+1}\right)$$
where $\varphi^2-\varphi-1=0$, and thus
$$I=\frac 52\int\frac{4v^2(2v-1)}{v+1}\left(\frac 1{v^2-\varphi v+1}+\frac 1{v^2+\frac v\varphi+1}\right)\text dv$$
Now we are in the realm of relatively simple integral terms.
