- #1

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(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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- #1

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(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?

- #2

HallsofIvy

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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

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Huh... Yes the denominators are the same.

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)?

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

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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

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where did you find this integral?

- #6

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My textbook in

Chapter 7 integration Techniques - Section 4 Partial Fractions

Chapter 7 integration Techniques - Section 4 Partial Fractions

- #7

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but it doesn't do the partial fraction thing

:(

Name of textbook and problem number please.

:(

Name of textbook and problem number please.

- #8

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(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

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

- #10

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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

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

- #12

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

- #13

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

- #14

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

- #15

SammyS

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Maybe you could come up with this....

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

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.

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- #17

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hmm thanksMaybe 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.

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