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

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In summary, the conversation discusses the process of evaluating the rational function (3x-5)/(x^2+5x+8) and the integral (x^2+2)/(x(x^2+5x+8)). The participants explore different methods, such as using partial fractions and completing the square, to solve the integral and discuss their approaches and difficulties. Finally, one of the participants shares a method involving multiplying the numerator by a constant and adding/subtracting a constant in order to integrate.
  • #1
GreenPrint
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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?
 
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  • #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)?
 
  • #3
HallsofIvy said:
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
 
  • #4
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
 
  • #5
where did you find this integral?
 
  • #6
My textbook in
Chapter 7 integration Techniques - Section 4 Partial Fractions
 
  • #7
but it doesn't do the partial fraction thing
:(

Name of textbook and problem number please.
 
  • #8
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

so I had

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
 
  • #9
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))
 
  • #10
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:
  • #11
that's what i get and I am trying to see if it's correct
 
  • #12
I'll post this in calculus and beyond sense this is now mostly calculus help
 
  • #13
OMG Sorry I got distracted by PHD! I'm back on the case now! (In the calc sections.)
 
  • #14
No, you know what? This integral BITES. Save it for your professor. Make him squirm. :P
 
  • #15
GreenPrint said:
...
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.

[itex]\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}[/itex]
[itex]\displaystyle =\frac{3}{2}\frac{2x+5}{x^2+5x+8}+\frac{3}{2}\frac{-\frac{10}{3}-5}{x^2+5x+8}[/itex]​

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.
 
  • #16
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
 
  • #17
SammyS said:
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.

[itex]\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}[/itex]
[itex]\displaystyle =\frac{3}{2}\frac{2x+5}{x^2+5x+8}+\frac{3}{2}\frac{-\frac{10}{3}-5}{x^2+5x+8}[/itex]​

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
 

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

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

The two expressions are equivalent because they both represent the same mathematical operation: dividing the numerator by the denominator. By using algebraic manipulation, we can see that the two expressions have the same simplified form.

How do we simplify the expression (3x-5)/(x^2+5x+8)?

To simplify this expression, we can factor the denominator and then cancel out any common factors in the numerator and denominator. In this case, the denominator can be factored into (x+4)(x+2), and we can cancel out the factor of (x+4) in both the numerator and denominator, leaving us with the simplified expression of (3x-5)/(x+2).

Why is it necessary to factor the denominator in this expression?

Factoring the denominator allows us to identify any common factors and simplify the expression. In this case, factoring the denominator allows us to cancel out the common factor of (x+4) and simplify the expression.

How do we simplify the expression (3(2x+5))/(2(x^2+5x+8))-25/(2(x^2+5x+8))?

We can factor the denominator of 2(x^2+5x+8) into 2(x+4)(x+2) and then cancel out the common factor of 2 in the numerator and denominator. This leaves us with the simplified expression of (3x+15)/(x+4)-25/(x+4). We can then combine the two fractions by finding a common denominator of (x+4) and simplifying the numerator, resulting in the final expression of (3x+15-25)/(x+4), or (3x-10)/(x+4).

Can we simplify the expression (3x-5)/(x^2+5x+8) further?

No, the expression (3x-5)/(x^2+5x+8) is already in its simplest form. We cannot factor or cancel out any more terms without changing the value of the expression.

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