Stricker42o2

2023-02-18

How to evaluate the integral $\int \frac{dx}{{x}^{3}+1}$?

Helen House

Partial fraction expansion of $\frac{1}{{x}^{3}+1}$:
-1 is a root of the denominator, therefore, $\left(x+1\right)$ is a factor:
$\frac{x+1}{x+1}\left(\frac{{x}^{2}-x+1}{\text{|}{x}^{3}+0{x}^{2}+0x+1}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\frac{-{x}^{3}-{x}^{2}}{\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}-{x}^{2}}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\frac{+{x}^{2}+x\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}}{\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}x+1}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}-x-1$
Verify the quotient's discriminant:
${b}^{2}-4\left(a\right)\left(c\right)=-{1}^{2}-4\left(1\right)\left(1\right)=-3$
The partial fractions are as follows because there are no real roots:
$\frac{1}{{x}^{3}+1}=\frac{Ax+B}{{x}^{2}-x+1}+\frac{C}{x+1}$
Multiply both sides by $\left({x}^{2}-x+1\right)\left(x+1\right)$:
$1=\left(Ax+B\right)\left(x+1\right)+C\left({x}^{2}-x+1\right)$
Make A and B disappear by letting x = -1:
$1=C\left({\left(-1\right)}^{2}--1+1\right)$
$C=\frac{1}{3}$
$1=\left(Ax+B\right)\left(x+1\right)+\frac{1}{3}\left({x}^{2}-x+1\right)$
Make A disappear by letting x = 0:
$1=B+\frac{1}{3}$
$B=\frac{2}{3}$
Let x = 1:
$1=\left(A+\frac{2}{3}\right)\left(2\right)+\frac{1}{3}$
$1=2A+\frac{5}{3}$
$A=-\frac{1}{3}$
$\frac{1}{{x}^{3}+1}=\frac{1}{3}\frac{1}{x+1}-\frac{1}{3}\frac{x-2}{{x}^{2}-x+1}$
Setting up the second term for ""u"" substitution:
$\frac{1}{{x}^{3}+1}=\frac{1}{3}\frac{1}{x+1}-\frac{1}{6}\frac{2x-4}{{x}^{2}-x+1}$
We want $2x-1$ in the numerator of the second term, therefore we much create a third term for the remaining -3:
$\frac{1}{{x}^{3}+1}=\frac{1}{3}\frac{1}{x+1}-\frac{1}{6}\frac{2x-1}{{x}^{2}-x+1}-\frac{1}{6}\frac{-3}{{x}^{2}-x+1}$
Now, the first two terms will integrate to natural logarithms and the last term will be a complete the square integral to become the inverse tangent:
$\frac{1}{{x}^{3}+1}=\frac{1}{3}\frac{1}{x+1}-\frac{1}{6}\frac{2x-1}{{x}^{2}-x+1}+\frac{1}{2}\frac{1}{{x}^{2}-x+1}$
Write each term as an integral:
$\int \frac{1}{{x}^{3}+1}dx=\frac{1}{3}\int \frac{1}{x+1}dx-\frac{1}{6}\int \frac{2x-1}{{x}^{2}-x+1}dx+\frac{1}{2}\int \frac{1}{{x}^{2}-x+1}dx$
$\int \frac{1}{{x}^{3}+1}dx=\frac{1}{3}\mathrm{ln}\left(x+1\right)-\frac{1}{6}\mathrm{ln}\left({x}^{2}-x+1\right)+\frac{\sqrt{3}}{3}{\mathrm{tan}}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$

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