# How do you integrate int (x+5)/((x+3)(x-2)(x-5))  using partial fractions?

Oct 19, 2016

$\frac{x + 5}{\left(x + 3\right) \left(x - 2\right) \left(x - 5\right)} = \frac{A}{x + 5} + \frac{B}{x - 2} + \frac{C}{x - 5}$
${x}^{2} \left(A + B + C\right) - x \left(7 A + 2 B + 6 C\right) + 10 A - 15 B - 6 C = x + 5$
A+B+C=0; 7A +2B+C=1; 10A-15B-6C=5
A=1/5; B=-1/5; C=0
$\int \left(\frac{x + 5}{\left(x + 3\right) \left(x - 2\right) \left(x - 5\right)}\right) = \int \frac{1}{5 \left(x + 3\right)} - \int \frac{1}{5 \left(x - 2\right)} = \frac{1}{5} \left(\ln \left(x + 3\right) - \ln \left(x - 2\right)\right) = \frac{1}{5} \ln \left(\frac{x + 3}{x - 2}\right)$

Oct 19, 2016

$= \frac{1}{20} \ln \left(x + 3\right) - \frac{7}{15} \ln \left(x - 2\right) + \frac{5}{12} \ln \left(x - 5\right) + C$

#### Explanation:

Let's do the partial fractions first
$\frac{x + 5}{\left(x + 3\right) \left(x - 2\right) \left(x - 5\right)} = \frac{A}{x + 3} + \frac{B}{x - 2} + \frac{C}{x - 5}$
$= \frac{A \left(x - 2\right) \left(x - 5\right) + B \left(x + 3\right) \left(x - 5\right) + C \left(x + 3\right) \left(x - 2\right)}{\left(x + 3\right) \left(x - 2\right) \left(x - 5\right)}$
so have the following
$x + 5 = A \left(x - 2\right) \left(x - 5\right) + B \left(x + 3\right) \left(x - 5\right) + C \left(x + 3\right) \left(x - 2\right)$
let $x = 2$ so $7 = 0 - 15 B + 0$ so $B = - \frac{7}{15}$
let $x = 5$ so $10 = 0 + 0 + 24 C$ => $C = \frac{5}{12}$
let $x = - 3$ so $2 = 40 A + 0 + 0$ => $A = \frac{1}{20}$
finally $\int \frac{\left(x + 5\right) \mathrm{dx}}{\left(x + 3\right) \left(x - 2\right) \left(x - 5\right)} = \frac{1}{20} \int \frac{\mathrm{dx}}{x + 3} - \frac{7}{15} \int \frac{\mathrm{dx}}{x - 2} + \frac{5}{12} \int \frac{\mathrm{dx}}{x - 5}$
$= \frac{1}{20} \ln \left(x + 3\right) - \frac{7}{15} \ln \left(x - 2\right) + \frac{5}{12} \ln \left(x - 5\right) + C$