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

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}


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