# How do you integrate int (x+5)/((x+3)(x-7)(x-5))  using partial fractions?

int (x+5)/((x+3)(x-7)(x-5) $\mathrm{dx}$

$= \frac{1}{40} \cdot \ln \left(x + 3\right) + \frac{3}{5} \cdot \ln \left(x - 7\right) - \frac{5}{8} \cdot \ln \left(x - 5\right) + K$

for a definite integral, a constant K must be added.

#### Explanation:

from the given $\int \frac{x + 5}{\left(x + 3\right) \left(x - 7\right) \left(x - 5\right)}$ $\mathrm{dx}$ =

$\int \frac{A}{x + 3}$$\mathrm{dx}$ + $\frac{B}{x - 7}$$\mathrm{dx}$ + $\frac{C}{x - 5}$*$\mathrm{dx}$

our equation becomes

(x+5)/((x+3)(x-7)(x-5)= $\frac{A}{x + 3} + \frac{B}{x - 7} + \frac{C}{x - 5}$

it follows;

(x+5)/((x+3)(x-7)(x-5)=

$\frac{A \left(x - 7\right) \left(x - 5\right) + B \left(x + 3\right) \left(x - 5\right) + C \left(x + 3\right) \left(x - 5\right)}{\left(x + 3\right) \left(x - 5\right) \left(x - 7\right)}$

by using only the numerators;

x+5=A(x^2-12x+35)+B(x^2-2x-15)+C(x^2-4x-21)

collecting the like terms

$x + 5 = \left(A + B + C\right) {x}^{2} + \left(- 12 A - 2 B - 4 C\right) x + \left(35 A - 15 B - 21 C\right)$

because the left side of the equation means

$0 \cdot {x}^{2} + 1 \cdot x + 5$

and the right side means

$\left(A + B + C\right) {x}^{2} + \left(- 12 A - 2 B - 4 C\right) x + \left(35 A - 15 B - 21 C\right)$

the equations are now formed;

$A + B + C = 0$ first equation

$- 12 A - 2 B - 4 C = 1$ second equation

$35 A - 15 B - 21 C = 5$ third equation

using your skills in solving 3 equations with 3 unknowns A, B, C

the values are $A = \frac{1}{40}$, $B = \frac{3}{5}$, and $C = - \frac{5}{8}$

Now, go back to the first line of the explanation to do the integration procedures.

$\int \frac{A}{x + 3}$$\mathrm{dx}$ + $\frac{B}{x - 7}$$\mathrm{dx}$ + $\frac{C}{x - 5}$*$\mathrm{dx}$

$= \int \frac{\frac{1}{40}}{x + 3}$$\mathrm{dx}$ + $\int \frac{\frac{3}{5}}{x - 7}$$\mathrm{dx}$ + $\int \frac{- \frac{5}{8}}{x - 5}$$\mathrm{dx}$

int (x+5)/((x+3)(x-7)(x-5) $\mathrm{dx}$
$= \frac{1}{40} \cdot \ln \left(x + 3\right) + \frac{3}{5} \cdot \ln \left(x - 7\right) - \frac{5}{8} \cdot \ln \left(x - 5\right) + K$