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

##### 1 Answer
Nov 6, 2015

$I = - \frac{1}{4} \ln \left({x}^{2} + 3\right) + \frac{1}{2 \sqrt{3}} \arctan \left(\frac{x}{\sqrt{3}}\right) + \frac{1}{2} \ln | x + 3 | + C$

#### Explanation:

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

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

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

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

$A + C = 0 \implies A = - C$
$B + 3 A = - 1 \implies B = - 1 + 3 C$
$3 B + 3 C = 3 \implies B + C = 1$

$- 1 + 3 C + C = 1 \implies 4 C = 2 \implies C = \frac{1}{2}$

$A = - \frac{1}{2}$

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

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

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

$I = - \frac{1}{4} \ln \left({x}^{2} + 3\right) + \frac{1}{2 \sqrt{3}} \arctan \left(\frac{x}{\sqrt{3}}\right) + \frac{1}{2} \ln | x + 3 | + C$