# How do you integrate f(x)=1/((x-2)(x-5)(x+3)) using partial fractions?

Apr 8, 2016

$\frac{1}{24} \ln | x - 5 | + \frac{1}{40} \ln | x + 3 | - \frac{1}{15} \ln | x - 2 | + c$

#### Explanation:

Since the factors on the denominator are linear then the numerators of the partial fractions will be constants , say A , B and C
$\Rightarrow \frac{1}{\left(x - 2\right) \left(x - 5\right) \left(x + 3\right)} = \frac{A}{x - 2} + \frac{B}{x - 5} + \frac{C}{x + 3}$

now, multiply through by (x-2)(x-5)(x+3)

so 1 = A(x-5)(x+3) + B(x-2)(x+3) + C(x-2)(x-5) ................................(1)

We now have to find the values of A , B and C. Note that if x = 2 , the terms with B and C will be zero. If x = 5 , the terms with A and C will be zero and if x = -3, the terms with A and B will be zero.
Making use of this fact , we obtain.

let x = 2 in (1) : 1 = -15A $\Rightarrow A = - \frac{1}{15}$

let x = 5 in (1) : 1 = 24B $\Rightarrow B = \frac{1}{24}$

let x = -3 in (1) : 1 = 40C $\Rightarrow C = \frac{1}{40}$

$\Rightarrow \frac{1}{\left(x - 2\right) \left(x - 5\right) \left(x + 3\right)} = \frac{- \frac{1}{15}}{x - 2} + \frac{\frac{1}{24}}{x - 5} + \frac{\frac{1}{40}}{x + 3}$

Integral now becomes.

$- \frac{1}{15} \int \frac{\mathrm{dx}}{x - 2} + \frac{1}{24} \int \frac{\mathrm{dx}}{x - 5} + \frac{1}{40} \int \frac{\mathrm{dx}}{x + 3}$

$= \frac{1}{24} \ln | x - 5 | + \frac{1}{40} \ln | x + 3 | - \frac{1}{15} \ln | x - 2 | + c$