# How do you integrate int (x-x^2)/((x+3)(x-6)(x+4))  using partial fractions?

Jan 18, 2016

$\int f \left(x\right) = - \frac{302}{5} \ln | x + 3 | - \frac{39}{5} \ln | x - 6 | + \frac{356}{5} \ln | x + 4 | + C$

#### Explanation:

First split the expression up into three elements, one for each of the bracketed terms in the denominator.

$f \left(x\right) = \frac{x - {x}^{2}}{\left(x + 3\right) \left(x - 6\right) \left(x + 4\right)}$
$f \left(x\right) = \frac{A}{x + 3} + \frac{B}{x - 6} + \frac{C}{x + 4}$

Restoring the denominator gives us
$f \left(x\right) = \frac{A \left(x - 6\right) \left(x + 4\right) + B \left(x + 3\right) \left(x + 4\right) + C \left(x + 3\right) \left(x - 6\right)}{\left(x + 3\right) \left(x - 6\right) \left(x + 4\right)}$
$f \left(x\right) = \frac{A {x}^{2} - 2 A x - 24 A + B {x}^{2} + 7 B x + 12 B + C {x}^{2} - 3 C x - 18 C}{\left(x + 3\right) \left(x - 6\right) \left(x + 4\right)}$

$= \frac{\left(A + B + C\right) {x}^{2} + \left(- 2 A + 7 B - 3 C\right) x + \left(- 24 A + 12 B - 18 C\right)}{\left(x + 3\right) \left(x - 6\right) \left(x + 4\right)}$

Each of the coefficients in the numerator must equal the coefficients in the original equation. Therefore
$A + B + C = - 1$
$- 2 A + 7 B - 3 C = 1$
$- 24 A + 12 B - 18 C = 0$

We can now solve these three equations to find A, B, and C.
Doubling the first equation and adding it to the second gives
$9 B - C = - 1$ or $C = 9 B + 1$

Multiplying the second equation by 12 and subtracting it from the third gives
$- 72 B + 18 C = - 12$ or $12 B + 3 C = - 2$

Substituting for C give $12 B + 3 \left(9 B + 1\right) = - 2$
$39 B = - 5$
$B = - \frac{39}{5}$
$C = 9 B + 1 = \frac{356}{5}$
$A = \frac{- 5 + 39 - 356}{5} = - \frac{302}{5}$

We can therefore write the original equation as
$F \left(x\right) = - \frac{302}{5 \left(x + 3\right)} - \frac{39}{5 \left(x - 6\right)} + \frac{356}{5 \left(x + 4\right)}$

$\int f \left(x\right) = - \frac{302}{5} \ln | x + 3 | - \frac{39}{5} \ln | x - 6 | + \frac{356}{5} \ln | x + 4 | + C$