# How do you find int (5x+11)/(x^2+2x-35) dx using partial fractions?

Apr 18, 2016

Do a partial fraction decomposition on $\frac{5 x + 11}{{x}^{2} + 2 x - 35}$ and simplify to get $2 \ln \left\mid x + 7 \right\mid + 3 \ln \left\mid x - 5 \right\mid + C$.

#### Explanation:

Begin by factoring the denominator to simplify the integral to:
$\int \frac{5 x + 11}{\left(x + 7\right) \left(x - 5\right)} \mathrm{dx}$

Because the denominator contains only linear factors, our partial fraction decomposition will be of the form:
$\frac{A}{x + 7} + \frac{B}{x - 5}$

We can now set up the decomposition:
$\frac{5 x + 11}{\left(x + 7\right) \left(x - 5\right)} = \frac{A}{x + 7} + \frac{B}{x - 5}$
$\frac{5 x + 11}{\left(x + 7\right) \left(x - 5\right)} = \frac{A \left(x - 5\right) + B \left(x + 7\right)}{\left(x + 7\right) \left(x - 5\right)}$

Equating the numerators and simplifying:
$5 x + 11 = A \left(x - 5\right) + B \left(x + 7\right)$
Set $x = 5$ to find the value of $B$:
$5 \left(5\right) + 11 = A \left(5 - 5\right) + B \left(5 + 7\right) \to 36 = 12 B \to B = 3$
Similarly, set $x = - 7$ to find $A$:
$5 \left(- 7\right) + 11 = A \left(- 7 - 5\right) + B \left(- 7 + 7\right) \to - 24 = - 12 A \to A = 2$

Our decomposition is therefore:
$\frac{5 x + 11}{\left(x + 7\right) \left(x - 5\right)} = \frac{2}{x + 7} + \frac{3}{x - 5}$

Putting this back into the integral and evaluating:
$\int \frac{5 x + 11}{\left(x + 7\right) \left(x - 5\right)} \mathrm{dx} = \int \frac{2}{x + 7} + \frac{3}{x - 5} \mathrm{dx}$
$\textcolor{w h i t e}{X X} = \int \frac{2}{x + 7} \mathrm{dx} + \int \frac{3}{x - 5} \mathrm{dx}$
$\textcolor{w h i t e}{X X} = 2 \int \frac{1}{x + 7} \mathrm{dx} + 3 \int \frac{1}{x - 5} \mathrm{dx}$
$\textcolor{w h i t e}{X X} = 2 \ln \left\mid x + 7 \right\mid + 3 \ln \left\mid x - 5 \right\mid + C$