# How do you integrate int (x+5)/((x+3)(x-2)(x-7))  using partial fractions?

Oct 5, 2017

The answer is $= \frac{1}{25} \ln \left(| x + 3 |\right) - \frac{7}{25} \ln \left(| x - 2 |\right) + \frac{6}{25} \ln \left(| x - 7 |\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{x + 5}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)} = \frac{A}{x + 3} + \frac{B}{x - 2} + \frac{C}{x - 7}$

$= \frac{A \left(x - 2\right) \left(x - 7\right) + B \left(x + 3\right) \left(x - 7\right) + C \left(x + 3\right) \left(x - 2\right)}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)}$

The denominators are the same, compare the numerators

$\left(x + 5\right) = \left(A \left(x - 2\right) \left(x - 7\right) + B \left(x + 3\right) \left(x - 7\right) + C \left(x + 3\right) \left(x - 2\right)\right)$

Let $x = - 3$, $\implies$, $2 = 50 A$, $A = \frac{1}{25}$

Let $x = 2$, $\implies$, $7 = - 25 B$, $B = - \frac{7}{25}$

Let $x = 7$, $\implies$, $12 = 50 C$, $C = \frac{6}{25}$

Therefore,

$\frac{x + 5}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)} = \frac{\frac{1}{25}}{x + 3} + \frac{- \frac{7}{25}}{x - 2} + \frac{\frac{6}{25}}{x - 7}$

So, the integral is

$\int \frac{\left(x + 5\right) \mathrm{dx}}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)} = \int \frac{\frac{1}{25} \mathrm{dx}}{x + 3} + \int \frac{- \frac{7}{25} \mathrm{dx}}{x - 2} + \int \frac{\frac{6}{25} \mathrm{dx}}{x - 7}$

$= \frac{1}{25} \ln \left(| x + 3 |\right) - \frac{7}{25} \ln \left(| x - 2 |\right) + \frac{6}{25} \ln \left(| x - 7 |\right) + C$