Integrating Rational Functions with Substitution: How to Solve Tricky Integrals

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Homework Help Overview

The discussion revolves around finding the integral of a rational function, specifically from 1/3 to 1/2 of the expression 1/x(x+1)(x-1)(x^2+1). Participants explore various substitution methods and the implications of partial fraction decomposition in the context of integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of substitution, with one suggesting u=x^4-1. There is uncertainty about the appropriate substitution and whether additional terms are needed in the numerator. Questions arise regarding the integration of the resulting expressions and the handling of limits.

Discussion Status

Several participants are actively engaging with the problem, offering suggestions and clarifications. There is a back-and-forth regarding the integration process, with some participants questioning assumptions and others providing insights into the decomposition of fractions. The discussion reflects a collaborative effort to navigate the complexities of the integral.

Contextual Notes

Participants note the importance of correctly applying limits after changing variables and the potential confusion that arises from improper notation. There is an emphasis on ensuring clarity in mathematical expressions throughout the discussion.

Shannabel
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Homework Statement


find the integral from 1/3 to 1/2 of 1/x(x+1)(x-1)(x^2+1)


Homework Equations





The Attempt at a Solution


expanded and simplified:
1/x^5-x

it gives a hint suggesting i use a substitution here, but i can't find an appropriate one.. Help?
 
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Welcome to PF, Shannabel! :smile:

Perhaps try for instance [itex]u=x^4-1[/itex]?
 
Are there suppose to be terms in the numerator besides 1? From what you wrote I would assume no but I wanted to make sure...
 
GreenPrint said:
Are there suppose to be terms in the numerator besides 1? From what you wrote I would assume no but I wanted to make sure...

nope, just a 1
 
Let me clean that up:
[tex]\int_{1/3}^{1/2} {1 \over x(x+1)(x-1)(x^2+1)} dx[/tex]

Click the Quote button to see what I did (it is called LaTeX). :wink:
 
I like Serena said:
Welcome to PF, Shannabel! :smile:

Perhaps try for instance [itex]u=x^4-1[/itex]?

i don't see how that will work.. since then du = 4x^3dx
 
Ya then I would stick with I like Serena's suggestion if you haven't learned partial fraction decomposition yet. Expanding is kind of tedious and there are some very common integrals than can be evaluated once you resolve that into partial fractions.
 
Shannabel said:
i don't see how that will work.. since then du = 4x^3dx

Yep. Keep going! :smile:
 
I like Serena said:
Yep. Keep going! :smile:

then i get
du/4x^4(u) ... but i can't integrate that?
 
  • #10
Shannabel said:
then i get
du/4x^4(u) ... but i can't integrate that?

Well, you still have x in your expression.
x should be eliminated... what would you get then?
 
  • #11
I like Serena said:
Well, you still have x in your expression.
x should be eliminated... what would you get then?

how do i eliminate the x?
 
  • #12
Shannabel said:
how do i eliminate the x?

Can you solve x from u=x4-1?
And substitute that?
 
  • #13
I like Serena said:
Can you solve x from u=x4-1?
And substitute that?

do you mean like x^4 = u+1?
 
  • #14
Shannabel said:
do you mean like x^4 = u+1?

Yep! :smile:
 
  • #15
then i get
du/u^2+u from 1/3 to 1/2
which is lnu^2 + ln u from 1/3 to 1/2
=2lnu+lnu
=3lnu from 1/3 to 1/2

u = x^4-1
so when x = (1/2), u = -15/16
and when x = (1/3), u = -80/81

so from above i have
3ln(-15/16)-3ln(-80/81)

where did i go wrong? lol
 
  • #16
I like Serena said:
Yep! :smile:

then i get
du/u^2+u from 1/3 to 1/2
which is lnu^2 + ln u from 1/3 to 1/2
=2lnu+lnu
=3lnu from 1/3 to 1/2

u = x^4-1
so when x = (1/2), u = -15/16
and when x = (1/3), u = -80/81

so from above i have
3ln(-15/16)-3ln(-80/81)

where did i go wrong? lol
 
  • #17
Not so fast! :eek:

Let's start with:

Shannabel said:
then i get
du/u^2+u from 1/3 to 1/2
which is lnu^2 + ln u from 1/3 to 1/2

And I'd really like some parentheses here, otherwise it becomes very confusing. :confused:

So you had: du/(u^2+u)

and now you're saying this integrates to: ln(u^2) + ln u

Let's check that...
Could you perhaps calculate the derivative of that last expression?
 
  • #18
I like Serena said:
Not so fast! :eek:

Let's start with:



And I'd really like some parentheses here, otherwise it becomes very confusing. :confused:

So you had: du/(u^2+u)

and now you're saying this integrates to: ln(u^2) + ln u

Let's check that...
Could you perhaps calculate the derivative of that last expression?

ohhh
i should have had
(du/u^2)+(du/u)
which integrates to
-1/u + ln u?
 
  • #19
Shannabel said:
ohhh
i should have had
(du/u^2)+(du/u)
which integrates to
-1/u + ln u?
Hmm, let's take a step back. :rolleyes:

You had:
Shannabel said:
then i get
du/4x^4(u) ... but i can't integrate that?
Shannabel said:
do you mean like x^4 = u+1?

So what do you get after you substitute (with the proper parentheses please)?
 
  • #20
I like Serena said:
Hmm, let's take a step back. :rolleyes:

You had:



So what do you get after you substitute (with the proper parentheses please)?

du/[4(u+1)(u)]?
 
  • #21
Shannabel said:
du/[4(u+1)(u)]?

Yes! :smile:

Do you know how to integrate that?
Since there is a sort of a trick to it...
 
  • #22
I like Serena said:
Yes! :smile:

Do you know how to integrate that?
Since there is a sort of a trick to it...

i can take the 1/4 out right?
and then i can separate into du/u^2 + du/u?
 
  • #23
Shannabel said:
i can take the 1/4 out right?

Yes! :smile:

Shannabel said:
and then i can separate into du/u^2 + du/u?

No. :frown:

(Sorry for the repetition! :wink:)
 
  • #24
I like Serena said:
Yes! :smile:



No. :frown:

(Sorry for the repetition! :wink:)


hmm.. okay then i think i am stumped :(
 
  • #25
Shannabel said:
hmm.. okay then i think i am stumped :(

The expression cannot be integrated straight away.
The easiest way to do this is a decomposition of fractions.

You have: 1/[(u+1)(u)].

You want to write this as: A/(u+1) + B/u.
Can you find an A and a B for which this is true (hint: combine it into 1 fraction)?
 
  • #26
I like Serena said:
The expression cannot be integrated straight away.
The easiest way to do this is a decomposition of fractions.

You have: 1/[(u+1)(u)].

You want to write this as: A/(u+1) + B/u.
Can you find an A and a B for which this is true (hint: combine it into 1 fraction)?
A=1 & B=-1

so (1/4) * du/u-du/(u+1)?
 
  • #27
Shannabel said:
so (1/4) * du/u-du/(u+1)?

Yep! :cool:

(Don't forget the parentheses and you should have reversed A and B! :wink:)
 
  • #28
I like Serena said:
Yep! :cool:

(Don't forget the parentheses and you should have reversed A and B! :wink:)


why are they reversed?
i had
a(u+1) + b(u) = 1
so that if u=0, a=1
and if u=-1, -b=1 so b=-1
?
 
  • #29
I like Serena said:
Yep! :cool:

(Don't forget the parentheses and you should have reversed A and B! :wink:)


nevermind, i wrote it down opposite from you :)

okay so now i have
1/4 * [ln(u)-ln(u+1)] between 1/3 and 1/2, right?
 
  • #30
Shannabel said:
why are they reversed?
i had
a(u+1) + b(u) = 1
so that if u=0, a=1
and if u=-1, -b=1 so b=-1
?

Well, your final expression was right.
And if you fill in the A and the B you find, you'll see that you wouldn't get your final expression.
As for the mistake you made, you should have:
a(u) + b(u+1) = 1
 

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