# How do you integrate int (5x - 10)/ (x^2+2x-35) dx using partial fractions?

Mar 22, 2018

The answer is $= \frac{15}{4} \ln \left(| x + 7 |\right) + \frac{5}{4} \ln \left(| x - 5 |\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{5 x - 10}{{x}^{2} + 2 x - 35} = \frac{5 x - 10}{\left(x + 7\right) \left(x - 5\right)} = \frac{A}{x + 7} + \frac{B}{x - 5}$

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

The denominators are the same, compare the numerators

$5 x - 10 = A \left(x - 5\right) + B \left(x + 7\right)$

Let $x = - 7$, $\implies$, $- 45 = - 12 A$, $\implies$, $A = \frac{45}{12} = \frac{15}{4}$

Let $x = 5$, $\implies$, $15 = 12 B$, $\implies$, $A = \frac{15}{12} = \frac{5}{4}$

Therefore,

$\frac{5 x - 10}{{x}^{2} + 2 x - 35} = \frac{\frac{15}{4}}{x + 7} + \frac{\frac{5}{4}}{x - 5}$

And finally,

$\int \frac{\left(5 x - 10\right) \mathrm{dx}}{{x}^{2} + 2 x - 35} = \int \frac{\frac{15}{4} \mathrm{dx}}{x + 7} + \int \frac{\frac{5}{4} \mathrm{dx}}{x - 5}$

$= \frac{15}{4} \ln \left(| x + 7 |\right) + \frac{5}{4} \ln \left(| x - 5 |\right) + C$