# How do you integrate int(x+1)/((x+5)(x+6)(x-1)) using partial fractions?

Oct 9, 2016

$\int \frac{x + 1}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)} \mathrm{dx} = \frac{2}{3} \ln | x + 5 | - \frac{5}{7} \ln | x + 6 | + \frac{1}{21} \ln | x - 1 | + c$

#### Explanation:

We first expand the given expression into partial fractions:

$\frac{x + 1}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)} \equiv \frac{A}{x + 5} + \frac{B}{x + 6} + \frac{C}{x - 1}$
$\frac{x + 1}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)} \equiv \frac{A \left(x + 6\right) \left(x - 1\right) + B \left(x + 5\right) \left(x - 1\right) + C \left(x + 5\right) \left(x + 6\right)}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)}$

And so. (x+1) -= A(x+6)(x-1) + B(x+5)(x-1) + C(x+5)(x+6))

Put $x = - 5 \implies - 5 + 1 = A \left(- 5 + 6\right) \left(- 5 - 1\right) + 0 + 0$
$\therefore \left(1\right) \left(- 6\right) A = - 4 \implies A = \frac{2}{3}$

Put $x = - 6 \implies - 6 + 1 = 0 + B \left(- 6 + 5\right) \left(- 6 - 1\right) + 0$
$\therefore \left(- 1\right) \left(- 7\right) A = - 5 \implies B = - \frac{5}{7}$

Put $x = 1 \implies 1 + 1 = 0 + 0 + C \left(1 + 5\right) \left(1 + 6\right)$
$\therefore \left(6\right) \left(7\right) C = 2 \implies C = \frac{1}{21}$

So the partial fraction decomposition is:
$\frac{x + 1}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)} \equiv \frac{2}{3 \left(x + 5\right)} - \frac{5}{7 \left(x + 6\right)} + \frac{1}{21 \left(x - 1\right)}$

We now want to integrate; so
$\int \frac{x + 1}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)} \mathrm{dx} = \int \left(\frac{2}{3 \left(x + 5\right)} - \frac{5}{7 \left(x + 6\right)} + \frac{1}{21 \left(x - 1\right)}\right) \mathrm{dx}$
$\int \frac{x + 1}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)} \mathrm{dx} = \frac{2}{3} \int \frac{1}{x + 5} \mathrm{dx} - \frac{5}{7} \int \frac{1}{x + 6} \mathrm{dx} + \frac{1}{21} \int \frac{1}{x - 1} \mathrm{dx}$

$\int \frac{x + 1}{\left(x + 5\right) \left(x + 6\right) \left(x - 1\right)} \mathrm{dx} = \frac{2}{3} \ln | x + 5 | - \frac{5}{7} \ln | x + 6 | + \frac{1}{21} \ln | x - 1 | + c$