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

Mar 6, 2017

Please see the explanation.

#### Explanation:

Write the equation for the expansion:

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

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

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

Make B and C disappear by letting x = 0:

$4 = 4 A$

$A = 1$

$4 = 1 \left({x}^{2} + 4\right) + B {x}^{2} + C x$

Let x = 1:

$4 = 5 + B + C$

$B + C = - 1$

Let x = -1:

$4 = 5 + B - C$

$B - C = - 1$

Clearly C = 0 and B = -1:

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

Modify the second integral for a variable substitution:

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

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