# Evaluate the integral int (1)/(x^3-1)dx

Evaluate the integral
$\int \frac{1}{{x}^{3}-1}dx$
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Monica Dennis
Evaluate the integral
$\int \frac{1}{{x}^{3}-1}dx=\int \frac{1}{\left(x-1\right)\left({x}^{2}+x+1\right)}dx\phantom{\rule{0ex}{0ex}}=\int \frac{A}{x-1}dx+\int \frac{Bx+C}{{x}^{2}+x+1}dx$
Solving $1=A\cdot \left({x}^{2}+x+1\right)+\left(Bx+C\right)\left(x-1\right)\phantom{\rule{0ex}{0ex}}1=A{x}^{2}+Ax+A+B{x}^{2}-Bx+Cx-C\phantom{\rule{0ex}{0ex}}1=\left(A+B\right){x}^{2}+\left(A-B+C\right)x+A-C\phantom{\rule{0ex}{0ex}}0=A+B⇒B=-A\phantom{\rule{0ex}{0ex}}0=A-B+C\phantom{\rule{0ex}{0ex}}1=A-C⇒C=A-1$

$=\frac{1}{3}\mathrm{ln}|x-1|-\frac{1}{6}\mathrm{ln}|w|-\frac{1}{2}\cdot \left(\frac{1}{\frac{\sqrt{3}}{2}}arctag\left(\frac{2}{\sqrt{3}}u\right)\right)+c\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{ln}|x-1|-\frac{1}{6}\mathrm{ln}|\left(x+\frac{1}{2}{\right)}^{2}+\frac{3}{4}|-\frac{\sqrt{3}}{3}arctag\left(\frac{2x+1}{\sqrt{3}}\right)+C$