# Why does (3x-5)/(x^2+5x+8) = (3(2x+5))/(2(x^2+5x+8))-25/(2(x^2+5x+8))?

I don't understand why
(3x-5)/(x^2+5x+8) = (3(2x+5))/(2(x^2+5x+8))-25/(2(x^2+5x+8))

and was hopping somebody could explain why it's not partial fractions or anything is it?

Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
Well, except for the "2"s on the right, the denominators are exactly the same aren't they?
So this is A/D= B/(2D)- C/(2D)= (1/2)(B/D- C/D)= (1/2)((B-C)/D).
Is "A" in this case the same as (1/2)(B- C)?

That is, what is (3/2)(2x+ 5)- (25/2)?

Well, except for the "2"s on the right, the denominators are exactly the same aren't they?
So this is A/D= B/(2D)- C/(2D)= (1/2)(B/D- C/D)= (1/2)((B-C)/D).
Is "A" in this case the same as (1/2)(B- C)?

That is, what is (3/2)(2x+ 5)- (25/2)?
Huh... Yes the denominators are the same.

A/D= B/(2D)- C/(2D)= (1/2)(B/D- C/D)= (1/2)((B-C)/D)

This is essentially splitting the rational function into two terms that are half the original over the same denominator and finding out the terms in the numerator for each half?

Yes A = (1/2)(B- C)

(3/2)(2x+ 5)- (25/2) = 3x - 5

huh interesting I never though of doing this...

The reason why I asked this was because I was trying to evaluate

integral (x^2+2)/(x(x^2+5x+8)) dx

and got to this point

1/4 ln|x| + c + 1/4 integral (3x-5)/(x^2+5x+8) dx

do you know of any other way of evaluating

integral (3x-5)/(x^2+5x+8) dx besides splitting (3x-5)/(x^2+5x+8) into (3(2x+5))/(2(x^2+5x+8))-25/(2(x^2+5x+8)) and taking two separate integrals because I don't think I would of come up with that on my own in like the middle of a test

I tried all sorts of things

x^2 + 5x + 8 can be expressed by completing the square as (x+5/2)^2+7/4

and you can than proceed to use a trig sub but every time i do it gets overly complicated

where did you find this integral?

My textbook in
Chapter 7 integration Techniques - Section 4 Partial Fractions

but it doesn't do the partial fraction thing
:(

Name of textbook and problem number please.

ya i know i got up to here

(x^2+2)/(x(x^2+5x+8)) = 1/(4x) + (3x-5)/(4(x^2+5x+8))

integral of the first term is just 1/4 ln(x) + c

1/4 ln(x) + c + 1/4 integral (3x-5)/(x^2+5x+8)

I then completed the square and tried to do a trig sub

This is what I got when I tried to do a trig sub

-3 ln|sqrt(7)/(2 sqrt((x+5/2)^2+7/4))| + (10*sqrt(7))/7*cot^(-1)((2x+3)/sqrt(7))

1/4 ln(|x|)-3/4 ln(|sqrt(7)/(2 sqrt((x+5/2)^2+7/4))|) + (10*sqrt(7))/28*cot^(-1)((2x+3)/sqrt(7))+c

Last edited:
that's what i get and im trying to see if it's correct

I'll post this in calculus and beyond sense this is now mostly calculus help

OMG Sorry I got distracted by PHD! I'm back on the case now! (In the calc sections.)

No, you know what? This integral BITES. Save it for your professor. Make him squirm. :P

SammyS
Staff Emeritus
Homework Helper
Gold Member
...
do you know of any other way of evaluating

integral (3x-5)/(x^2+5x+8) dx besides splitting (3x-5)/(x^2+5x+8) into (3(2x+5))/(2(x^2+5x+8))-25/(2(x^2+5x+8)) and taking two separate integrals because I don't think I would of come up with that on my own in like the middle of a test
Maybe you could come up with this.

The derivative of the denom. is 2x+5.

Just multiply the numerator by 2/3 and add ( and subtract) the needed constant term -- to the numerator also.

$\displaystyle \frac{3}{2}\frac{(2/3)(3x-5)}{x^2+5x+8}=\frac{3}{2}\frac{2x-\frac{10}{3}+5 -5}{x^2+5x+8}$
$\displaystyle =\frac{3}{2}\frac{2x+5}{x^2+5x+8}+\frac{3}{2}\frac{-\frac{10}{3}-5}{x^2+5x+8}$​

The first term gives a log when integrated.

Clean up the second term, of course the numer. is -25, then complete the square in the denom. & integrate to get an arctan.

I solved it see my thread in calculus and beyond section i decided to post it there because now that i understood the basic math i needed help in calculus... it did suck but i was able to solve it lol

Maybe you could come up with this.

The derivative of the denom. is 2x+5.

Just multiply the numerator by 2/3 and add ( and subtract) the needed constant term -- to the numerator also.

$\displaystyle \frac{3}{2}\frac{(2/3)(3x-5)}{x^2+5x+8}=\frac{3}{2}\frac{2x-\frac{10}{3}+5 -5}{x^2+5x+8}$
$\displaystyle =\frac{3}{2}\frac{2x+5}{x^2+5x+8}+\frac{3}{2}\frac{-\frac{10}{3}-5}{x^2+5x+8}$​

The first term gives a log when integrated.

Clean up the second term, of course the numer. is -25, then complete the square in the denom. & integrate to get an arctan.
hmm thanks