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

Jan 8, 2017

The answer is =x+1/2ln(∣x+2∣)+5/2ln(∣x-2∣)+C

#### Explanation:

Since the degree of the numerator is not less than the degree of the denominator, perform a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} + 3 x$$\textcolor{w h i t e}{a a a a a a a a}$∣${x}^{2} - 4$

$\textcolor{w h i t e}{a a a a}$${x}^{2}$$\textcolor{w h i t e}{a a a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$∣$1$

$\textcolor{w h i t e}{a a a a}$$0 + 3 x$$\textcolor{w h i t e}{a a a}$$+ 4$

Therefore,

By factorising the denominator,

$\frac{{x}^{2} + 3 x}{{x}^{2} - 4} = 1 + \frac{3 x + 4}{{x}^{2} - 4} = 1 + \frac{3 x + 4}{\left(x + 2\right) \left(x - 2\right)}$

Now, we perform the partial fraction decomposition

$\frac{3 x + 4}{\left(x + 2\right) \left(x - 2\right)} = \frac{A}{x + 2} + \frac{B}{x - 2}$

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

So,

$3 x + 4 = A \left(x - 2\right) + B \left(x + 2\right)$

Let $x = 2$, $\implies$, $10 = 4 B$, $\implies$, $B = \frac{5}{2}$

Let $x = - 2$, $\implies$, $- 2 = - 4 A$, $\implies$, $A = \frac{1}{2}$

Therefore,

$\frac{{x}^{2} + 3 x}{{x}^{2} - 4} = 1 + \frac{\frac{1}{2}}{x + 2} + \frac{\frac{5}{2}}{x - 2}$

So,

$\int \frac{\left({x}^{2} + 3 x\right) \mathrm{dx}}{{x}^{2} - 4} = \int 1 \mathrm{dx} + \left(\frac{1}{2}\right) \int \frac{\mathrm{dx}}{x + 2} + \left(\frac{5}{2}\right) \int \frac{\mathrm{dx}}{x - 2}$

=x+1/2ln(∣x+2∣)+5/2ln(∣x-2∣)+C