# How do you integrate int (2x+1)/((x+4)(x-1)(x+7))  using partial fractions?

Feb 26, 2016

$\frac{7}{15} \ln | x + 4 | + \frac{3}{40} \ln | x - 1 | - \frac{13}{24} \ln | x + 7 | + c$

#### Explanation:

Since the factors on the denominator of the rational function are linear, then the numerators of the partial fractions will be constants, say A , B and C.

hence $\frac{2 x + 1}{\left(x + 4\right) \left(x - 1\right) \left(x + 7\right)} = \frac{A}{x + 4} + \frac{B}{x - 1} + \frac{C}{x + 7}$

multiply through by (x+4)(x-1)(x+7) to obtain:

2x+1 = A(x-1)(x+7) + B(x+4)(x+7) + C(x+4)(x-1) ................(1)

The aim now is to find the value of A , B and C. Note that if x = -4 then the terms with B and C will be zero. If x = 1 the terms with A and C will be zero and if x =-7 the terms with A and B will be zero.
This is the starting point in finding A , B and C.

let x = -4 in (1) : -7 = -15A $\Rightarrow A = \frac{7}{15}$

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

let x =-7 in (1) : -13 = 24C $\Rightarrow C = - \frac{13}{24}$

$\frac{2 x + 1}{\left(x + 4\right) \left(x - 1\right) \left(x + 7\right)} = \frac{\frac{7}{15}}{x + 4} + \frac{\frac{3}{40}}{x - 1} - \frac{\frac{13}{24}}{x + 7}$

integral becomes:

$\int \frac{\frac{7}{15}}{x + 4} \mathrm{dx} + \int \frac{\frac{3}{40}}{x - 1} \mathrm{dx} - \int \frac{\frac{13}{24}}{x + 7} \mathrm{dx}$

$= \frac{7}{15} \ln | x + 4 | + \frac{3}{40} \ln | x - 1 | - \frac{13}{24} \ln | x + 7 | + c$