# Thread: Can someone explain how this intergral gets rewritten

1. ## Can someone explain how this intergral gets rewritten

Studying for my maths unit for my part 66 license and my example question has a bit that is massively confusing me and I hope someone with better maths skills than me can help [so that when i do my exam i have a chance of getting this right]

The question is this: $\int\frac{x-4}{x^2-2x-3}$

And the first part of the given solution is:

$3\int x-4/(x^2-2x-3)$

Rearrange intergrand $\frac{x-4}{x^2-2x-3}$ into form $\frac{2x-2}{2(x^2-2x-3)} - \frac{3}{x^2-2x-3}$

However the example doesnt explain how that rearannged part is obtained and I've been staring at it for the last 2 hours trying to work out how

Can anyone explain [preferably idiot-proof] how this rearranging bit works?

Many Thanks

2. ## Re: Can someone explain how this intergral gets rewritten

Originally Posted by Saz
Studying for my maths unit for my part 66 license and my example question has a bit that is massively confusing me and I hope someone with better maths skills than me can help [so that when i do my exam i have a chance of getting this right]
The question is this: $\int\frac{x-4}{x^2-2x-3}$
Look at this webpage.
Ben sure to click the show steps tab.

3. ## Re: Can someone explain how this intergral gets rewritten

But that still doesnt show how the intergral bit gets rewritten and im not going to have internet access in the middle of an exam. It just says the same thing the example says: rearrange a to get b without showing how you get from a to b

4. ## Re: Can someone explain how this intergral gets rewritten

Originally Posted by Saz
But that still doesnt show how the intergral bit gets rewritten and im not going to have internet access in the middle of an exam. It just says the same thing the example says: rearrange a to get b without showing how you get from a to b
The show steps tab explains it well.

5. ## Re: Can someone explain how this intergral gets rewritten

?? The show steps tab says:

"rewrite intergrand $\int \frac{x-4}{x^2-2x-3as}$ as $\int \frac{2x-3}{2(x^2-2x-3)} - \int \frac{3}{x^2-2x-3}$"

...so how does that bit work? how do i get that first intergral to split into the 2 seperate intergrals? Like i said i need to know how that bit works so when i get a similar question i got a chance at solving it right

6. ## Re: Can someone explain how this intergral gets rewritten

Originally Posted by Saz
?? The show steps tab says:

"rewrite intergrand $\int \frac{x-4}{x^2-2x-3as}$ as $\int \frac{2x-2}{2(x^2-2x-3)} - \int \frac{3}{x^2-2x-3}$"

...so how does that bit work? how do i get that first intergral to split into the 2 seperate intergrals? Like i said i need to know how that bit works so when i get a similar question i got a chance at solving it right
You mean that you have trouble with simple algebra.
$\frac{2x-2}{2(x^2-2x-2)} - \frac{3}{x^2-2x-3}=\frac{(2x-2)-2(3)}{2(x^2-2x-3)}=\frac{x-4}{(x^2-2x-3)}$

7. ## Re: Can someone explain how this intergral gets rewritten

Originally Posted by Plato
You mean that you have trouble with simple algebra.
$\frac{2x-2}{2(x^2-2x-2)} - \frac{3}{x^2-2x-3}=\frac{(2x-2)-2(3)}{2(x^2-2x-3)}=\frac{x-4}{(x^2-2x-3)}$
I'm going to say yes, its been many years since ive done algebra and maths is not my strong point. i'm still not getting it how it works in reverse [i.e go from one fraction to two] is it something to do with partial fractions? I'm not quite sure. It's bad I can fix aircraft but not do maths lol. Sorry for being a pain

8. ## Re: Can someone explain how this intergral gets rewritten

no, this is not anything to do with partial fractions.

if we are going to integrate something like (ax+b)/(cx2 + dx + e)

it would be a lot easier if the top was the derivative of the bottom. because then we would have log|(something)| when we integrate.

well the derivative of x2 - 2x - 3 is 2x - 2. so 2x - 2 is what we "wish" we had in the numerator. well we can make x-4 look "more" like 2x - 2, by multiplying the top and bottom of our fraction by 2, and taking the 2 in the denominator outside the integral:

$\int \frac{x - 4}{x^2 - 2x - 3} dx = \frac{1}{2}\int \frac{2x - 8}{x^2 - 2x - 3} dx$.

now we have 2x - 8 in the top, which we split into the sum 2x - 2 - 6 (so now we have 2 integrals, with the same denominator):

$=\frac{1}{2}\int \frac{2x - 2}{x^2 - 2x - 3} dx - \frac{1}{2}\int \frac{6}{x^2 - 2x - 3} dx$

in the second integral, we can bring the 6 outside the integral, where it cancels with the 1/2 to make 3.

9. ## Re: Can someone explain how this intergral gets rewritten

Originally Posted by Deveno
no, this is not anything to do with partial fractions.

if we are going to integrate something like (ax+b)/(cx2 + dx + e)

it would be a lot easier if the top was the derivative of the bottom. because then we would have log|(something)| when we integrate.

well the derivative of x2 - 2x - 3 is 2x - 2. so 2x - 2 is what we "wish" we had in the numerator. well we can make x-4 look "more" like 2x - 2, by multiplying the top and bottom of our fraction by 2, and taking the 2 in the denominator outside the integral:

$\int \frac{x - 4}{x^2 - 2x - 3} dx = \frac{1}{2}\int \frac{2x - 8}{x^2 - 2x - 3} dx$.

now we have 2x - 8 in the top, which we split into the sum 2x - 2 - 6 (so now we have 2 integrals, with the same denominator):

$=\frac{1}{2}\int \frac{2x - 2}{x^2 - 2x - 3} dx - \frac{1}{2}\int \frac{6}{x^2 - 2x - 3} dx$

in the second integral, we can bring the 6 outside the integral, where it cancels with the 1/2 to make 3.
Ah that makes more sense now! Can't say I ever remember doing this type of intergration at GCSE and A-Level [although it has been a while and my memory is bad]

Thank you very much You've been a big help