# How do you integrate [((3x^2)+3x+12) / ((x-5)(x^2+9))]dx using partial fractions?

##### 1 Answer
Nov 12, 2016

Please see the explanation

#### Explanation:

Expand the fraction:

$\frac{3 {x}^{2} + 3 x + 12}{\left(x - 5\right) \left({x}^{2} + 9\right)} = \frac{A}{x - 5} + \frac{B x + C}{{x}^{2} + 9}$

Multiply both sides by the left side denominator:

$3 {x}^{2} + 3 x + 12 = A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x - 5\right)$

Let x = 5 to make B and C disapper

$3 {\left(5\right)}^{2} + 3 \left(5\right) + 12 = A \left({5}^{2} + 9\right)$

$102 = A \left(34\right)$

$A = 3$

Substitute 3 for A:

$3 {x}^{2} + 3 x + 12 = 3 \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x - 5\right)$

Let x = 0 to make B disappear:

$12 = 3 \left(9\right) + C \left(- 5\right)$

$- 15 = - 5 C$

$C = 3$

Substitute 3 for C:

$3 {x}^{2} + 3 x + 12 = 3 \left({x}^{2} + 9\right) + \left(B x + 3\right) \left(x - 5\right)$

Let x = 1:

$3 + 3 + 12 = 3 \left(1 + 9\right) + \left(B + 3\right) \left(1 - 5\right)$

$18 = 30 - 4 B - 12$

$B = 0$

$\int \left(\frac{3 {x}^{2} + 3 x + 12}{\left(x - 5\right) \left({x}^{2} + 9\right)}\right) \mathrm{dx} = 3 \int \frac{1}{x - 5} \mathrm{dx} + 3 \int \frac{1}{{x}^{2} + 5} \mathrm{dx}$

$\int \left(\frac{3 {x}^{2} + 3 x + 12}{\left(x - 5\right) \left({x}^{2} + 9\right)}\right) \mathrm{dx} = 3 \ln | x - 5 | + 3 \frac{\sqrt{5}}{5} {\tan}^{-} 1 \left(\left(\sqrt{5}\right) \frac{x}{5}\right) + C$