# How do you integrate (3x^2+2x)/((x+2)(x^2+4)) using partial fractions?

Oct 28, 2016

 int(3x^2+2x)/((x+2)(x^2+4)) dx= ln(x+2)+ ln(x^2+4)−arctan(x/2)+C

#### Explanation:

The partial fraction decomposition will be of the form:

$\frac{3 {x}^{2} + 2 x}{\left(x + 2\right) \left({x}^{2} + 4\right)} \equiv \frac{A}{x + 2} + \frac{B x + C}{{x}^{2} + 4}$
$\therefore \frac{3 {x}^{2} + 2 x}{\left(x + 2\right) \left({x}^{2} + 4\right)} = \frac{A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x + 2\right)}{\left(x + 2\right) \left({x}^{2} + 4\right)}$
$\therefore 3 {x}^{2} + 2 x = A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x + 2\right)$

We now need to find the three coefficients, $A$,$B$ and $C$:

Put $x = - 2 \implies 12 - 4 = A \left(4 + 4\right) + 0$
$\therefore 8 A = 8 \implies A = 1$

Equating coefficients of ${x}^{2} \implies 3 = A + B$
$\therefore B = 3 - 1 = 2$

Equating coefficients of constants $\implies 0 = 4 A + 2 C$
$\therefore 2 C = - 4 \implies C = - 2$

So, we have:
$\frac{3 {x}^{2} + 2 x}{\left(x + 2\right) \left({x}^{2} + 4\right)} = \frac{1}{x + 2} + \frac{2 x - 2}{{x}^{2} + 4}$

And, therefore,
$\int \frac{3 {x}^{2} + 2 x}{\left(x + 2\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \int \frac{1}{x + 2} + \frac{2 x - 2}{{x}^{2} + 4} \mathrm{dx}$
$\therefore \int \frac{3 {x}^{2} + 2 x}{\left(x + 2\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \int \frac{1}{x + 2} \mathrm{dx} + \int \frac{2 x - 2}{{x}^{2} + 4} \mathrm{dx}$

 :. int(3x^2+2x)/((x+2)(x^2+4)) dx= ln(x+2)+ ln(x^2+4)−arctan(x/2)+C

I have omitted the derivation of the second interval, as the question is about partial fractions and not integration by substation, you can use http://www.integral-calculator.com/ to get a full step by step.