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

Apr 25, 2017

$\int \setminus \frac{2 x + 1}{\left(x - 2\right) \left({x}^{2} + 4\right)} \setminus \mathrm{dx} = \frac{5}{8} \ln | x - 2 | - \frac{5}{16} \ln | {x}^{2} + 4 | - \frac{3}{8} \arctan \left(\frac{x}{2}\right) + C$

#### Explanation:

Let:

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

We can decompose the integrand into partial fraction; as:

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

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

Put $x = 2 \implies 5 = 8 A \implies A = \frac{5}{8}$

Comparing Coefficients:

$\text{Coeff} \left({x}^{2}\right) \implies 0 = A + B$
$\text{Coeff} \left({x}^{0}\right) \implies 1 = 4 A - 2 C$

And so:

$B = - \frac{5}{8}$
$2 C = \frac{20}{8} - 1 \implies C = \frac{3}{4}$

Hence the partial fraction decomposition gives us:

$I = \int \setminus \frac{\frac{5}{8}}{x - 2} + \frac{\left(- \frac{5}{8}\right) x + \left(\frac{6}{8}\right)}{{x}^{2} + 4} \setminus \mathrm{dx}$
$\setminus \setminus = \frac{5}{8} \int \setminus \frac{1}{x - 2} \setminus \mathrm{dx} - \frac{1}{8} \int \setminus \frac{5 x - 6}{{x}^{2} + 4} \setminus \mathrm{dx} \setminus \setminus \ldots \left(\star\right)$

The first integral we can just evaluate directly,

$\int \setminus \frac{1}{x - 2} \setminus \mathrm{dx} = \ln | x - 2 |$

We can deal with the second integral by splitting into two fractions:

$\int \setminus \frac{5 x - 6}{{x}^{2} + 4} \setminus \mathrm{dx} = \int \setminus \frac{5 x}{{x}^{2} + 4} - \frac{6}{{x}^{2} + 4} \setminus \mathrm{dx}$
$\text{ } = \frac{5}{2} \int \setminus \frac{2 x}{{x}^{2} + 4} \setminus \mathrm{dx} - \int \setminus \frac{6}{{x}^{2} + 4} \setminus \mathrm{dx}$
$\text{ } = \frac{5}{2} \ln | {x}^{2} + 4 | - \frac{6}{2} \arctan \left(\frac{x}{2}\right)$

Substituting these three integral results into $\left(\star\right)$ we get:

$I = \frac{5}{8} \ln | x - 2 | - \frac{1}{8} \left\{\frac{5}{2} \ln | {x}^{2} + 4 | - 3 \arctan \left(\frac{x}{2}\right)\right\} + C$
$\setminus \setminus = \frac{5}{8} \ln | x - 2 | - \frac{5}{16} \ln | {x}^{2} + 4 | - \frac{3}{8} \arctan \left(\frac{x}{2}\right) + C$