# How do you integrate int (2x) / (x^3 + 1) using partial fractions?

Dec 21, 2016

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

#### Explanation:

The denominator factorises as $\left(x + 1\right) \left({x}^{2} - x + 1\right)$. So the integrand is
$\frac{- \frac{2}{3}}{x + 1} + \frac{A x + B}{{x}^{2} - x + 1}$ where the $\frac{2}{3}$ comes from the cover-up rule: $\frac{\left(2\right) \left(- 1\right)}{{\left(- 1\right)}^{2} - \left(- 1\right) + 1}$ and $A$ and $B$ are to be found by equating coefficients of suitable powers of $x$ or otherwise:
$2 x \equiv \left(- \frac{2}{3}\right) \left({x}^{2} - x + 1\right) + \left(A x + B\right) \left(x + 1\right)$ which, upon re-grouping in powers of $x$ gives:
$0 {x}^{2} + 2 x + 0 \equiv \left(- \frac{2}{3} + A\right) {x}^{2} + \left(\frac{2}{3} + A + B\right) x - \frac{2}{3} + B$.
So $A = \frac{2}{3}$, $B = \frac{2}{3}$

So the integral is:
$\int \frac{2}{3} \left(- \frac{1}{x + 1} + \frac{1}{2} \frac{2 x - 1 + 3}{{x}^{2} - x + 1}\right) \mathrm{dx}$
$= \frac{2}{3} \left(- \ln | x + 1 |\right) + \frac{1}{3} \ln | {x}^{2} - x + 1 | + \int \frac{1}{{\left(x - \frac{1}{2}\right)}^{2} + \frac{3}{4}} \mathrm{dx}$
$= \frac{2}{3} \left(- \ln | x + 1 |\right) + \frac{1}{3} \ln | {x}^{2} - x + 1 | + \left(\frac{2}{\sqrt{3}}\right) {\tan}^{-} 1 \left(\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C$
$= \frac{2}{3} \left(- \ln | x + 1 |\right) + \frac{1}{3} \ln | {x}^{2} - x + 1 | + \left(\frac{2}{\sqrt{3}}\right) {\tan}^{-} 1 \left(\frac{2 x - 1}{\sqrt{3}}\right) + C$

Notice the trick of splitting the numerator $x + 1$ into the multiple of the derivative of the denominator plus a correction constant (here $+ 3$) in anticipation of using the rule about integrating $\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) | + C$