Question

evaluate the following integral using partial fraction decomposition.
int\frac{9}{(x - 5)(x + 4)}dx
int\frac{9}{(x - 5)(x + 4)}dx=square
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Decompose into partial - fractions

Let $\frac{9}{(x - 5)(x + 4)}=\frac{A}{x - 5}+\frac{B}{x + 4}$. Then $9=A(x + 4)+B(x - 5)$.
Set $x = 5$, we get $9=A(5 + 4)+B(5 - 5)$, so $9 = 9A$, and $A = 1$.
Set $x=-4$, we get $9=A(-4 + 4)+B(-4 - 5)$, so $9=-9B$, and $B=-1$.
So $\frac{9}{(x - 5)(x + 4)}=\frac{1}{x - 5}-\frac{1}{x + 4}$.

Step2: Integrate the decomposed fractions

$\int\frac{9}{(x - 5)(x + 4)}dx=\int(\frac{1}{x - 5}-\frac{1}{x + 4})dx$.
Using the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have $\int\frac{1}{x - 5}dx-\int\frac{1}{x + 4}dx=\ln|x - 5|-\ln|x + 4|+C$.
By the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, it can be written as $\ln|\frac{x - 5}{x + 4}|+C$.

Answer:

$\ln|\frac{x - 5}{x + 4}|+C$