# How do you integrate int 1 / ((x+5)^2 (x-1) )  using partial fractions?

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#### Explanation:

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Sep 20, 2016

$\int \frac{1}{{\left(x + 5\right)}^{2} \left(x + 1\right)} \mathrm{dx} = - \frac{1}{36} \ln | x + 5 | + \frac{1}{6 \left(x + 5\right)} + \frac{1}{36} \ln | x - 1 | + C$

#### Explanation:

Since the denominator has been factorized as a combination of linear factors, one repeated, we may write it in partial fraction form as :

$\frac{1}{{\left(x + 5\right)}^{2} \left(x + 1\right)} = \frac{A}{x + 5} + \frac{B}{x + 5} ^ 2 + \frac{C}{x - 1}$

$= \frac{A \left(x + 5\right) \left(x - 1\right) + B \left(x - 1\right) + C {\left(x + 5\right)}^{2}}{{\left(x + 5\right)}^{2} \left(x - 1\right)}$

$= \frac{\left(A + C\right) {x}^{2} + \left(4 A + B + 10 C\right) x + \left(- 5 A - B + 25 C\right)}{{\left(x + 5\right)}^{2} \left(x - 1\right)}$

By comparing terms in the numerator, we now see that :

$A + C = 0$
$4 A + B + 10 C = 0$
$- 5 A - B + 25 C = 1$

Solving this system of linear equations yields :

$A = - \frac{1}{36} , B = - \frac{1}{6} , C = \frac{1}{36}$

Hence the original integral may be written and solved as follows :

$\int \frac{1}{{\left(x + 5\right)}^{2} \left(x + 1\right)} \mathrm{dx} = - \frac{1}{36} \int \frac{1}{x + 5} \mathrm{dx} - \frac{1}{6} \int \frac{1}{x + 5} ^ 2 \mathrm{dx} + \frac{1}{36} \int \frac{1}{x - 1} \mathrm{dx}$

$= - \frac{1}{36} \ln | x + 5 | + \frac{1}{6 \left(x + 5\right)} + \frac{1}{36} \ln | x - 1 | + C$

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