# How do you integrate 8 / [ ( x + 2 ) ( x^2 + 4 ) ]  using partial fractions?

Apr 15, 2017

#### Explanation:

Decompose: $\frac{8}{\left(x + 2\right) \left({x}^{2} + 4\right)}$

$\frac{8}{\left(x + 2\right) \left({x}^{2} + 4\right)} = \frac{A}{x + 2} + \frac{B x}{{x}^{2} + 4} + \frac{C}{{x}^{2} + 4}$

Multiply both sides by $\left(x + 2\right) \left({x}^{2} + 4\right)$:

$8 = A \left({x}^{2} + 4\right) + B x \left(x + 2\right) + C \left(x + 2\right)$

Let $x = - 2$:

$8 = A \left({\left(- 2\right)}^{2} + 4\right) + B \left(- 2\right) \left(- 2 + 2\right) + C \left(- 2 + 2\right)$

$8 = A \left(8\right) + B \left(- 2\right) \left(0\right) + C \left(0\right)$

$A = 1$

$4 - {x}^{2} = B x \left(x + 2\right) + C \left(x + 2\right)$

Let $x = 0$

$4 - {0}^{2} = B \left(0\right) \left(0 + 2\right) + C \left(0 + 2\right)$

$4 = 2 C$

$C = 2$

$4 - {x}^{2} = B x \left(x + 2\right) + 2 x + 4$

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

$B = - 1$

$\int \frac{8}{\left(x + 2\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \int \frac{1}{x + 2} \mathrm{dx} - \int \frac{x}{{x}^{2} + 4} \mathrm{dx} + \int \frac{2}{{x}^{2} + 4} \mathrm{dx}$

The first integral is the natural logarithm:

$\int \frac{8}{\left(x + 2\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \ln | x + 2 | - \int \frac{x}{{x}^{2} + 4} \mathrm{dx} + \int \frac{2}{{x}^{2} + 4} \mathrm{dx}$

Multiple the second integral by $\frac{2}{2}$ and it, too, becomes a natural logarithm but without an absolute value, because the argument can never be negative:

$\int \frac{8}{\left(x + 2\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \ln | x + 2 | - \frac{1}{2} \ln \left({x}^{2} + 4\right) + \int \frac{2}{{x}^{2} + 4} \mathrm{dx}$

The last integral is our old friend the inverse tangent:

$\int \frac{8}{\left(x + 2\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \ln | x + 2 | - \frac{1}{2} \ln \left({x}^{2} + 4\right) + {\tan}^{-} 1 \left(\frac{x}{2}\right) + C$