# How do you find int (x-1)/((x^3+1))dx using partial fractions?

Nov 15, 2015

$I = - \frac{2}{3} \ln | x + 1 | + \frac{1}{3} \ln | {x}^{2} - x + 1 | + C$

#### Explanation:

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

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

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

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

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

$A + B = 0$
$- A + B + C = 1$
$A + C = - 1$

$2 B + C = 1$
$B + 2 C = 0$

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

$C = 1 - 2 B = - \frac{1}{3}$

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

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

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

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

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

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

$I = - \frac{2}{3} \ln | x + 1 | + \frac{1}{3} \ln | {x}^{2} - x + 1 | + C$