<s>[INST] The right way of proving LHS = RHS I'm currently brushing up on my math before heading back to school for my bachelor's and am practicing on some questions.
Given the following equation :
$$\frac{x^2 - 3x}{x^2 - 9} = 1-\frac{3x-9}{x^2-9}.$$
and that I have to prove that the LHS = RHS; How would I go about doing so?
What I've done:
I've simply brought $(3x-9)/(x^2-9)$ over to the LHS, resulting in the equation $(x^2 - 3x)/(x^2-9) + (3x-9)/(x^2-9) = 1$ , which gives me $(x^2-9)/(x^2-9) = 1$.(LHS = RHS)
However, I do not think that what I've done is the correct method of proving. Should I have instead transformed the LHS into RHS without bringing anything from LHS to RHS and vice versa?
Furthermore , after which I'm integrating $(x^2-3x)/(x^2-9)$ by doing the following:
What I first did was factorise the denominator into $(x+3)(x-3)$.
I then equated $(x^2 - 3x) = A(x + 3) + B(x - 3)$.
Given that $x = 3 \implies A = 0$ and given that $x = -3 \implies B = -3$.
This would then give me the following equation to integrate:
$[ (-3)/(x+3) + 0(x-3) ]dx$
which led me to the answer $-3ln|x+3| + C$.
However, when I checked my answer, the correct answer should be 
$$x -3ln|x+3| + C.$$
Where did the missing $x$ come from?
Any assistance would be greatly appreciated
Thanks!
 [/INST] First, to show the desired equality, we have: $$\frac{x^2-3x}{x^2-9}=\frac{x^2-9-3x+9}{x^2-9}=\frac{x^2-9}{x^2-9}-\frac{3x-9}{x^2-9}=1-\frac{3x-9}{x^2-9}.$$
Now, assuming that you are trying to find the integral of $\dfrac{x^2-3x}{x^2-9}$, we have: 
\begin{align}
\int{\dfrac{x^2-3x}{x^2-9}dx}&=\int\left({1-\dfrac{3x-9}{x^2-9}}\right)dx \\
&=\int{\left(1-3\cdot\frac{x-3}{(x-3)(x+3)}\right)dx} \\
&=\int{dx}-3\int{\frac{1}{x+3}dx} \\
&=x-3\ln|x+3|+C.
\end{align}
</s>