How do you evaluate the integral of #int 1/[(x^3)-1] dx#?

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Answer 1 · 2018-07-11T19:40:55

#\color{red}{\int \frac{1}{x^3-1}\ dx}=\color{blue}{1/6\ln|\frac{x^2-2x+1}{x^2+x+1}|-1/\sqrt3\tan^{-1}(\frac{2x+1}{\sqrt3})+C}#

#\int \frac{1}{x^3-1}\ dx#

#=\int \frac{1}{(x-1)(x^2+x+1)}\ dx#

#=\int(\frac{1}{3(x-1)}-\frac{x+2}{3(x^2+x+1)})\ dx#

#=\int(\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)})\ dx#

#=1/3\int\frac{1}{x-1}\ dx-1/6\int \frac{2x+1}{x^2+x+1}\ dx-1/2\int \frac{dx}{x^2+x+1}#

#=1/3\int\frac{d(x-1)}{x-1}-1/6\int \frac{d(x^2+x+1)}{x^2+x+1}-1/2\int \frac{dx}{(x+1/2)^2+3/4}#

#=1/3\ln|x-1|-1/6\ln|x^2+x+1|-1/2\int \frac{d(x+1/2)}{(x+1/2)^2+(\sqrt3/2)^2}#

#=1/6\ln|\frac{(x-1)^2}{x^2+x+1}|-1/2\cdot (1/{\sqrt3/2})\tan^{-1}(\frac{x+1/2}{\sqrt3/2})+C#

#=1/6\ln|\frac{x^2-2x+1}{x^2+x+1}|-1/\sqrt3\tan^{-1}(\frac{2x+1}{\sqrt3})+C#