Evaluate the following integral:

Solution:

$I=\int \frac{x^{3}-1}{x^{3}+x} d x=\int 1-\frac{x+1}{x^{3}+x} d x$

$=\int 1 \mathrm{dx}-\int \frac{\mathrm{x}+1}{\mathrm{x}^{3}+\mathrm{x}} \mathrm{dx}$

$\frac{\mathrm{x}+1}{\mathrm{x}\left(\mathrm{x}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+1}$

$x+1=A\left(x^{2}+1\right)+(B x+C)(x)$

Equating constants

$A=1$

Equating coefficients of $x$

$1=\mathrm{C}$

Equating coefficients of $x^{2}$

$0=A+B$

$B=-1$

$I=-\int \frac{d x}{x}-\int \frac{-x+1 d x}{x^{2}+1}+\int d x$

$I=-\int \frac{d x}{x}+\int \frac{x d x}{x^{2}+1}-\int \frac{d x}{x^{2}+1}+\int d x$

$=-\log |\mathrm{x}|+\frac{1}{2} \log \left|\mathrm{x}^{2}+1\right|-\tan ^{-1} \mathrm{x}+\mathrm{x}+\mathrm{c}$

$I=x-\log |x|+\frac{1}{2} \log \left|x^{2}+1\right|-\tan ^{-1} x+c$