# How do you evaluate the integral int dx/(x^3+1)?

Jan 21, 2017

Use partial fraction expansion and then integrate

#### Explanation:

Partial fraction expansion of $\frac{1}{{x}^{3} + 1}$:

-1 is a root of the denominator, therefore, $\left(x + 1\right)$ is a factor:

color(white)((x +1)/color(black)(x+1)(color(black)((x^2 - x + 1)/("|"x^3 + 0x^2 + 0x + 1))
color(white)(" ")(-x^3 - x^2)/(" "-x^2
$\textcolor{w h i t e}{\text{ ")(+x^2 + x" ")/(" } x + 1}$
$\textcolor{w h i t e}{\text{ }} - x - 1$

Check the discriminant of the quotient:

${b}^{2} - 4 \left(a\right) \left(c\right) = - {1}^{2} - 4 \left(1\right) \left(1\right) = - 3$

There are no real roots, therefore, the partial fractions are:

$\frac{1}{{x}^{3} + 1} = \frac{A x + B}{{x}^{2} - x + 1} + \frac{C}{x + 1}$

Multiply both sides by $\left({x}^{2} - x + 1\right) \left(x + 1\right)$:

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

Make A and B disappear by letting x = -1:

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

$C = \frac{1}{3}$

$1 = \left(A x + B\right) \left(x + 1\right) + \frac{1}{3} \left({x}^{2} - x + 1\right)$

Make A disappear by letting x = 0:

$1 = B + \frac{1}{3}$

$B = \frac{2}{3}$

Let x = 1:

$1 = \left(A + \frac{2}{3}\right) \left(2\right) + \frac{1}{3}$
$1 = 2 A + \frac{5}{3}$
$A = - \frac{1}{3}$

$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{3} \frac{x - 2}{{x}^{2} - x + 1}$

Setting up the second term for "u" substitution:

$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{6} \frac{2 x - 4}{{x}^{2} - x + 1}$

We want $2 x - 1$ in the numerator of the second term, therefore we much create a third term for the remaining -3:

$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{6} \frac{2 x - 1}{{x}^{2} - x + 1} - \frac{1}{6} \frac{- 3}{{x}^{2} - x + 1}$

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:

$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{6} \frac{2 x - 1}{{x}^{2} - x + 1} + \frac{1}{2} \frac{1}{{x}^{2} - x + 1}$

Write each term as an integral:

$\int \frac{1}{{x}^{3} + 1} \mathrm{dx} = \frac{1}{3} \int \frac{1}{x + 1} \mathrm{dx} - \frac{1}{6} \int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx} + \frac{1}{2} \int \frac{1}{{x}^{2} - x + 1} \mathrm{dx}$

$\int \frac{1}{{x}^{3} + 1} \mathrm{dx} = \frac{1}{3} \ln \left(x + 1\right) - \frac{1}{6} \ln \left({x}^{2} - x + 1\right) + \frac{\sqrt{3}}{3} {\tan}^{-} 1 \left(\frac{2 x - 1}{\sqrt{3}}\right) + C$