Evaluate the following integral:
$\int \frac{x^{3}-1}{x^{3}+x} d x$
$I=\int \frac{x^{3}-1}{x^{3}+x} d x=\int 1-\frac{x+1}{x^{3}+x} d x$
$=\int 1 \mathrm{dx}-\int \frac{\mathrm{x}+1}{\mathrm{x}^{3}+\mathrm{x}} \mathrm{dx}$
$\frac{\mathrm{x}+1}{\mathrm{x}\left(\mathrm{x}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+1}$
$x+1=A\left(x^{2}+1\right)+(B x+C)(x)$
Equating constants
$A=1$
Equating coefficients of $x$
$1=\mathrm{C}$
Equating coefficients of $x^{2}$
$0=A+B$
$B=-1$
$I=-\int \frac{d x}{x}-\int \frac{-x+1 d x}{x^{2}+1}+\int d x$
$I=-\int \frac{d x}{x}+\int \frac{x d x}{x^{2}+1}-\int \frac{d x}{x^{2}+1}+\int d x$
$=-\log |\mathrm{x}|+\frac{1}{2} \log \left|\mathrm{x}^{2}+1\right|-\tan ^{-1} \mathrm{x}+\mathrm{x}+\mathrm{c}$
$I=x-\log |x|+\frac{1}{2} \log \left|x^{2}+1\right|-\tan ^{-1} x+c$
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