\(\begin{aligned}
& I=\int\left(\frac{x^3-1}{x^3+x}\right) d x=\int\left(1-\frac{x+1}{x^3+x}\right) d x \\
& \Rightarrow \int 1 \cdot d x-\int \frac{(x+1)}{x^3+x} d x=x-\int \frac{(x+1)}{x\left(x^2+1\right)} d x
\end{aligned}\)
Now, \(\frac{x+1}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{x^2+1}\)
\(\Rightarrow \quad(x+1)=A\left(x^2+1\right)+(B x+C) x\)
[using partial fractions]
\(\Rightarrow \quad(x+1)=(A+B) x^2+C x+A\)
On comparing \(A+B=0, C=1, A=1\)
\(\begin{array}{lc}
\Rightarrow & B=-1 \\
\therefore & I=x-\int \frac{1}{x} d x-\int \frac{(1-x)}{x^2+1} d x \\
\Rightarrow & I=x-\log |x|-\int \frac{1}{x^2+1} d x+\frac{1}{2} \int \frac{2 x}{x^2+1} d x \\
\Rightarrow & I=x-\log |x|-\tan ^{-1} x+\frac{1}{2} \log \left(x^2+1\right)+C
\end{array}\)