# How do you use partial fraction decomposition to decompose the fraction to integrate (7)/(x^2+13x+40)?

Mar 7, 2018

The integral equals $\frac{7}{3} \ln | \frac{x + 5}{x + 8} | + C$

#### Explanation:

We wish to find factors in the denominator. The trick is to find two numbers that multiply to $40$ and add to $13$. Clearly these will be $8$ and $5$.

$I = \int \frac{7}{\left(x + 5\right) \left(x + 8\right)} \mathrm{dx}$

Now we can decompose in partial fractions.

$\frac{A}{x + 5} + \frac{B}{x + 8} = \frac{7}{\left(x + 5\right) \left(x + 8\right)}$

$A \left(x + 8\right) + B \left(x + 5\right) = 7$

$A x + 8 A + B x + 5 B = 7$

$\left(A + B\right) x + \left(8 A + 5 B\right) = 7$

Now we have a system of equations.

$\left\{\begin{matrix}A + B = 0 \\ 8 A + 5 B = 7\end{matrix}\right.$

Substituting the first equation into the second we see that

$8 A + 5 \left(- A\right) = 7$

$3 A = 7$

$A = \frac{7}{3}$

Now clearly $B = - \frac{7}{3}$ because $A + B = 0$. The integral becomes:

$I = \int \frac{7}{3 \left(x + 5\right)} - \frac{7}{3 \left(x + 8\right)} \mathrm{dx}$

$I = \frac{7}{3} \ln | x + 5 | - \frac{7}{3} \ln | x + 8 | + C$

$I = \frac{7}{3} \ln | \frac{x + 5}{x + 8} | + C$

Hopefully this helps!