Put $x^4=t$, so that $4 x^3 d x=d t$
$\begin{aligned}
&\therefore \quad \mathrm{I}=\frac{1}{4} \int \frac{d t}{t(t-1)} ; \text { Let } \frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1} \\
&\Rightarrow \quad 1=\mathrm{A}(t-1)+\mathrm{B} t \quad \ldots(i)
\end{aligned}$
Put $t=0,1$ in (i), we get; $\mathrm{A}=-1 \& \mathrm{~B}=1$
$\begin{aligned}
&\therefore \mathrm{I}=\frac{1}{4} \int\left(\frac{-1}{t}+\frac{1}{t-1}\right) d t \\
&=\frac{1}{4} \log \left|\frac{t-1}{t}\right|+\mathrm{C}=\frac{1}{4} \log \left|\frac{x^4-1}{x^4}\right|+\mathrm{C}
\end{aligned}$