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