$I=\int \large\frac{dx}{x(x^3+8)}$
Mutiply and divide by $x^2$
$I=\large\frac {x^2dx}{x^3(x^3+8)}$
Let $x^3+8=t=>x^3=t=8$
$3x^2dx=dt$
$x^2dx=dt/3$
Therefore $I=\frac{1}{3} \int \frac{dt}{(t-8)(t)}=\frac{1}{3}\int \bigg[\frac{A}{t-8}+\frac{B}{t}\bigg]dt$
$A(t)+B(t-8)=1$
Put $t=8 \qquad 8A=1=>A=1/8$
Put $ t=0 \qquad =>B=-1/8$
Therefore $I=\large\frac{1}{3} \int\frac{dt}{8(t-8)}-\frac{1}{24}\int \frac{dt}{t}$
$=\frac{1}{3}\bigg[\large\frac{1}{8} log |t-8|-\frac{1}{8} log |t|\bigg]+c$
$=\large\frac{1}{3}\bigg[\frac{1}{8}log \frac{|t-8|}{|t|}\bigg]+c$
$=\large\frac{1}{24}log \frac{1}{8}log \frac{|t-8|}{|t|}+c$
Substituting for t
$\large\frac{1}{24}log \frac{|x^3+8-8|}{|x^3+8|}+c$
$\large\frac{1}{24}log \frac{|x^3|}{|x^3+8|}+c$