# How do you integrate int (x+4) / [(x-1)(x^2+4)] using partial fractions?

Nov 23, 2016

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

#### Explanation:

The Partial fraction decomposition of the integrand will be of the form;

$\frac{x + 4}{\left(x - 1\right) \left({x}^{2} + 4\right)} \equiv \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 4}$
$\therefore \frac{x + 4}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 1\right)}{\left(x - 1\right) \left({x}^{2} + 4\right)}$
$\therefore \left(x + 4\right) \equiv A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 1\right)$

Put $x = 1 \implies 5 = A \left(1 + 4\right) + 0$
$\therefore 5 A = 5 \implies A = 1$

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

Compare coefficients of ${x}^{2} \implies 0 = A + B$
$\therefore B = - 1$

Hence:
$\frac{x + 4}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{1}{x - 1} - \frac{x}{{x}^{2} + 4}$

And so,

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