Hi, hope someone can help with this. Any working or explanation would be greatly appreciated. I really want to understand why/what I need to do/think. Thank you.
Find the indefinite integral of dx/x^2 -9
Cheers
I assume you mean this
$\displaystyle \int \frac{dx}{x^2-9}$
The first thing to do is to factor the denominator and then use partial fraction decomp
$\displaystyle \frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}$
Clearing the fractions we get
$\displaystyle 1=A(x+3)+B(x-3)$
if we let $\displaystyle x=3$ we get
$\displaystyle 1=A(6) \iff A=\frac{1}{6}$
if we let $\displaystyle x=-3$ we get
$\displaystyle 1=B(-6) \iff B=-\frac{1}{6}$
Now we have the integrals
$\displaystyle \int \frac{dx}{x^2-9}=\int \frac{\frac{1}{6}}{(x-3)}dx+\int \frac{-\frac{1}{6}}{(x+3)}dx$
The latter are both u subs for the denominator