# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{2 \mathrm{x}}{\mathrm{x}^{3}-1} \mathrm{dx}$

Solution:

$I=\int \frac{2 x}{x^{3}-1} d x=\int \frac{2 x}{(x-1)\left(x^{2}+x+1\right)} d x$

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

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

$=(A+B) x^{2}+(A-B+C) x+(A-C)$

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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