# How do you integrate (2x+3)/(x^2-9) using partial fractions?

Apr 17, 2018

$\int \frac{2 x + 3}{{x}^{2} - 9} \mathrm{dx} = \frac{\ln \left\mid x + 3 \right\mid + 3 \ln \left\mid x - 3 \right\mid}{2} + C$

#### Explanation:

Since the numerator is in a lower degree than the denominator, we can proceed to the next step.

$\int \frac{2 x + 3}{{x}^{2} - 9} \mathrm{dx}$ Factor

$\int \frac{2 x + 3}{\left(x + 3\right) \left(x - 3\right)} \mathrm{dx}$

We can rewrite the rational function in the form:

$\frac{A}{x + 3} + \frac{B}{x - 3}$

$\implies \frac{A}{x + 3} \cdot \frac{x - 3}{x - 3} + \frac{B}{x - 3} \cdot \frac{x + 3}{x + 3}$

$\implies \frac{A \left(x - 3\right)}{\left(x + 3\right) \left(x - 3\right)} + \frac{B \left(x + 3\right)}{\left(x - 3\right) \left(x + 3\right)}$

$\implies \frac{A \left(x - 3\right) + B \left(x + 3\right)}{\left(x - 3\right) \left(x + 3\right)}$

$\implies \frac{A x - 3 A + B x + 3 B}{\left(x - 3\right) \left(x + 3\right)}$

$\implies \frac{A x + B x + 3 B - 3 A}{\left(x - 3\right) \left(x + 3\right)}$

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

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

We let:

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

$3 \left(B - A\right) = 3$

$\implies A + B = 2$

$\implies - A + B = 1$ Add the equations together.

$\implies 2 B = 3$

$\implies B = \frac{3}{2}$

$\implies A = \frac{1}{2}$

We substitute this in the original form to get:

$\int \frac{1}{2} \cdot \frac{1}{x + 3} + \frac{3}{2} \cdot \frac{1}{x - 3} \mathrm{dx}$

$\implies \int \frac{1}{2} \cdot \frac{1}{x + 3} \mathrm{dx} + \int \frac{3}{2} \cdot \frac{1}{x - 3} \mathrm{dx}$

$\implies \frac{1}{2} \int \frac{1}{x + 3} \mathrm{dx} + \frac{3}{2} \int \frac{1}{x - 3} \mathrm{dx}$

Remember that:

$\int \frac{1}{x + n} \mathrm{dx} = \ln \left\mid x + n \right\mid$

$\implies \frac{1}{2} \cdot \ln \left\mid x + 3 \right\mid + \frac{3}{2} \ln \left\mid x - 3 \right\mid$

$\implies \ln \frac{\left\mid x + 3 \right\mid}{2} + \frac{3 \ln \left\mid x - 3 \right\mid}{2}$

$\implies \frac{\ln \left\mid x + 3 \right\mid + 3 \ln \left\mid x - 3 \right\mid}{2}$ Do you $C$ why this is incomplete?

$\implies \frac{\ln \left\mid x + 3 \right\mid + 3 \ln \left\mid x - 3 \right\mid}{2} + C$