# How do you integrate int (2x+1)/((x+3)(x-2)(x-3))  using partial fractions?

Apr 19, 2016

$- \frac{1}{6} \ln \left\mid x + 3 \right\mid - \ln \left\mid x - 2 \right\mid + \frac{7}{6} \ln \left\mid x - 3 \right\mid + C$

#### Explanation:

First, focus on writing the fraction in terms of partial fractions:

$\frac{2 x + 1}{\left(x + 3\right) \left(x - 2\right) \left(x - 3\right)} = \frac{A}{x + 3} + \frac{B}{x - 2} + \frac{C}{x - 3}$

Multiply both sides by $\left(x + 3\right) \left(x - 2\right) \left(x - 3\right)$:

$2 x + 1 = A \left(x - 2\right) \left(x - 3\right) + B \left(x + 3\right) \left(x - 3\right) + C \left(x + 3\right) \left(x - 2\right)$

Expand:

$2 x + 1 = A \left({x}^{2} - 5 x + 6\right) + B \left({x}^{2} - 9\right) + C \left({x}^{2} + x - 6\right)$

Sort by degree:

$2 x + 1 = {x}^{2} \left(A + B + C\right) + x \left(- 5 A + C\right) + \left(6 A - 9 B - 6 C\right)$

Giving the system:

$\left\{\begin{matrix}A + B + C = 0 \\ - 5 A + C = 2 \\ 6 A - 9 B - 6 C = 1\end{matrix}\right.$

Solving which gives:

$\left\{\begin{matrix}A = - \frac{1}{6} \\ B = - 1 \\ C = \frac{7}{6}\end{matrix}\right.$

Thus, we know that

$\int \frac{2 x + 1}{\left(x + 3\right) \left(x - 2\right) \left(x - 3\right)} \mathrm{dx} = \int - \frac{1}{6 \left(x + 3\right)} - \frac{1}{x - 2} + \frac{7}{6 \left(x - 3\right)} \mathrm{dx}$

Splitting up the integral and refactoring constants:

$= - \frac{1}{6} \int \frac{1}{x + 3} \mathrm{dx} - \int \frac{1}{x - 2} \mathrm{dx} + \frac{7}{6} \int \frac{1}{x - 3} \mathrm{dx}$

$= - \frac{1}{6} \ln \left\mid x + 3 \right\mid - \ln \left\mid x - 2 \right\mid + \frac{7}{6} \ln \left\mid x - 3 \right\mid + C$