I'm struggling figuring out how to break apart the fraction and what to use in the numerator (Ax+B, A+B+C?)
help!
INTEGRAL 4x^2 +36 / x(x^2 -9)
thanks
$\displaystyle \frac{4x^2+36}{x(x+3)(x-3)} = \frac{A}{x} + \frac{B}{x+3} + \frac{C}{x-3}
$
$\displaystyle 4x^2 + 36 = A(x+3)(x-3) + Bx(x-3) + Cx(x+3)
$
let $\displaystyle x = 0$ ...
$\displaystyle 36 = -9A$ ... $\displaystyle A = -4$
let $\displaystyle x = -3$ ...
$\displaystyle 72 = 18B$ ... $\displaystyle B = 4$
let $\displaystyle x = 3$ ...
$\displaystyle 72 = 18C$ ... $\displaystyle C = 4$
$\displaystyle 4 \int -\frac{1}{x} + \frac{1}{x+3} + \frac{1}{x-3} \, dx $
The particular letters you use are up to you!
This is really a topic from algebra, but it's usually only touched on in calculus, with the assumption (usually wrong, in my experience) that the student is already familiar with the technique, which can be frustrating!
For a review with worked examples, look through a few online lessons.