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


  • $\begingroup$ If you manage to prove that $a+b=1$ then you have managed to prove that $a=1-b$. So what you did originally is okay. You might add that the equality is only true if $x^2-9\neq0$ or equivalently $x\neq3\wedge x\neq-3$. $\endgroup$
    – drhab
    Nov 17 '15 at 11:34
  • 1
    $\begingroup$ You left out the $1$ from $1-\frac{3x-9}{x^2-9}$ when you wrote the equation to integrate. $\endgroup$
    – Nicholas
    Nov 17 '15 at 11:41
  • $\begingroup$ You can solve LHS and RHS and reach to a point also saying that both are equal since both have equal values. If you start with LHS, the expression simplifies to x/(x+3), and if you do the same thing with RHS, its still the same thing i.e. x/(x+3). So, we can see that both have equal values/expressions and therefore we can conclude that LHS = RHS. $\endgroup$ Feb 6 '20 at 0:10

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}

  • $\begingroup$ This is also a good way of showing LHS = RHS. $\endgroup$ Feb 6 '20 at 0:11

I've simply brought $(3x−9)/(x^2−9)$ over to the LHS,...

By so doing you have already assumed the equality of the two sides. When proving an alleged equality, one has access to only one side, not both - the other side should pop out from the proof.

As for the rest, its integration via partial fractions, which I think you were on the right track.


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