How do you use partial fraction decomposition to decompose the fraction to integrate (2x)/(1-x^3)?

Apr 14, 2018

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

Explanation:

Factors of $1 - {x}^{3}$ are $\left(1 - x\right) \left(1 + x + {x}^{2}\right)$, hence partial fractions of $\frac{2 x}{1 - {x}^{3}}$ are of the form

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

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

Comparing coefficients of constant term, $x$ and ${x}^{2}$ in numerator

$A + C = 0$, $A + B - C = 2$ and $A - B = 0$

solving these we get $A = B = \frac{2}{3}$, $C = - \frac{2}{3}$ and hence

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

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

Now $\int \frac{2}{3 \left(1 - x\right)} \mathrm{dx} = \frac{2}{3} \int \frac{1}{1 - x} \mathrm{dx} = - \frac{2}{3} \ln | x - 1 |$

and $\int \frac{2 x - 2}{3 \left(1 + x + {x}^{2}\right)} \mathrm{dx}$

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

= $\frac{1}{3} \ln | 1 + x + {x}^{2} | - 3 \left[\frac{2}{\sqrt{3}} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right)\right]$

For former observe $\frac{d}{\mathrm{dx}} \left(1 + x + {x}^{2}\right) = 2 x + 1$ and for latter, we can have

$\int \frac{3}{1 + x + {x}^{2}} \mathrm{dx} = 3 \int \frac{1}{{\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}} \mathrm{dx} = 3 \int \frac{1}{{\left(x + \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} \mathrm{dx}$

= $3 \left[\frac{2}{\sqrt{3}} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right)\right]$

Hence $\int \frac{2 x}{1 - {x}^{3}} \mathrm{dx} = - \frac{2}{3} \ln | x - 1 | + \frac{1}{3} \ln | 1 + x + {x}^{2} | - 3 \left[\frac{2}{\sqrt{3}} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right)\right]$