# How do you integrate int(x+1)/((x-3)(x-1)(x+4)) using partial fractions?

Feb 9, 2016

$\frac{2}{7} \ln | x - 3 | - \frac{1}{5} \ln | x - 1 | - \frac{3}{35} \ln | x + 4 | + c$

#### Explanation:

since the factors on the denominator are linear , the numerators will be constants.

$\Rightarrow \frac{x + 1}{\left(x - 3\right) \left(x - 1\right) \left(x + 4\right)} = \frac{A}{x - 3} + \frac{B}{x - 1} + \frac{C}{x + 4}$

multiply through by (x-3)(x-1)(x+4)

x+ 1 = A(x-1)(x+4) + B(x-3)(x+4) + C(x-3)(x-1).................(1)

now require to find values of A , B and C. Note that if x=1 , the terms with A and C will be zero. If x =3 , the terms with B and C will be zero and if x = -4 , the terms with A and B will be zero.
This is the starting point in finding values for A , B and C.

let x = 1 in (1) : 2 = - 10B → B =$- \frac{1}{5}$

let x = 3 in (1) : 4 = 14A → A $= \frac{2}{7}$

let x = -4 in (1) : -3 = 35C → C$= - \frac{3}{35}$

$\int \frac{\left(x + 1\right)}{\left(x - 3\right) \left(x - 1\right) \left(x + 4\right)} \mathrm{dx} = \int \left(\frac{\frac{2}{7}}{x - 3} - \frac{\frac{1}{5}}{x - 1} - \frac{\frac{3}{35}}{x + 4}\right) \mathrm{dx}$

$= \frac{2}{7} \ln | x - 3 | - \frac{1}{5} \ln | x - 1 | - \frac{3}{35} \ln | x + 4 | + c$

where c, is the constant of integration.