0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.