# How do you integrate (7x-2)/((x-3)^2(x+1)) using partial fractions?

Nov 3, 2017

Write the partial fractions equation:

$\frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} = \frac{A}{x + 1} + \frac{B}{x - 3} + \frac{C}{x - 3} ^ 2$

Multiply both sides by ${\left(x - 3\right)}^{2} \left(x + 1\right)$:

$7 x - 2 = A {\left(x - 3\right)}^{2} + B \left(x - 3\right) \left(x + 1\right) + C \left(x + 1\right)$

Let $x = - 1$

$7 \left(- 1\right) - 2 = A {\left(- 1 - 3\right)}^{2} + B \left(- 1 - 3\right) \left(- 1 + 1\right) + C \left(- 1 + 1\right)$

$- 9 = 16 A$

$A = - \frac{9}{16}$

$7 x - 2 = - \frac{9}{16} {\left(x - 3\right)}^{2} + B \left(x - 3\right) \left(x + 1\right) + C \left(x + 1\right)$

Let $x = 3$

$7 \left(3\right) - 2 = - \frac{9}{16} {\left(3 - 3\right)}^{2} + B \left(3 - 3\right) \left(3 + 1\right) + C \left(3 + 1\right)$

#C = 19/4

$7 x - 2 = - \frac{9}{16} {\left(x - 3\right)}^{2} + B \left(x - 3\right) \left(x + 1\right) + \frac{19}{4} \left(x + 1\right)$

Let $x = 0$

$7 \left(0\right) - 2 = - \frac{9}{16} {\left(0 - 3\right)}^{2} + B \left(0 - 3\right) \left(0 + 1\right) + \frac{19}{4} \left(0 + 1\right)$

$- 2 = - \frac{81}{16} - 3 B + \frac{19}{4} \left(0 + 1\right)$

$B = \frac{9}{16}$

The partial fractions are:

$\frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} = - \frac{9}{16} \frac{1}{x + 1} + \frac{9}{16} \frac{1}{x - 3} + \frac{19}{4} \frac{1}{x - 3} ^ 2$

Integrate:

$\int \frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} \mathrm{dx} = - \frac{9}{16} \int \frac{1}{x + 1} \mathrm{dx} + \frac{9}{16} \int \frac{1}{x - 3} \mathrm{dx} + \frac{19}{4} \int \frac{1}{x - 3} ^ 2 \mathrm{dx}$

The integrals are trivial:

$\int \frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} \mathrm{dx} = - \frac{9}{16} \ln t | x + 1 | + \frac{9}{16} \ln | x - 3 | - \frac{19}{4} \frac{1}{x - 3} + C$