Integrating Rational Functions with Substitution: How to Solve Tricky Integrals

[Shannabel]

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?

 

[I like Serena]

Welcome to PF, Shannabel!

Perhaps try for instance u=x^4-1?

 

[GreenPrint]

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...

 

[Shannabel]

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

 

[I like Serena]

Let me clean that up:
\int_{1/3}^{1/2} {1 \over x(x+1)(x-1)(x^2+1)} dx

Click the Quote button to see what I did (it is called LaTeX).

 

[Shannabel]

Welcome to PF, Shannabel!

Perhaps try for instance u=x^4-1?

i don't see how that will work.. since then du = 4x^3dx

 

[GreenPrint]

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.

 

[I like Serena]

i don't see how that will work.. since then du = 4x^3dx

Yep. Keep going!

 

[Shannabel]

Yep. Keep going!

then i get
du/4x^4(u) ... but i can't integrate that?

 

[I like Serena]

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?

 

[Shannabel]

Well, you still have x in your expression.
x should be eliminated... what would you get then?

how do i eliminate the x?

 

[I like Serena]

how do i eliminate the x?

Can you solve x from u=x⁴-1?
And substitute that?

 

[Shannabel]

Can you solve x from u=x⁴-1?
And substitute that?

do you mean like x^4 = u+1?

 

[I like Serena]

do you mean like x^4 = u+1?

Yep!

 

[Shannabel]

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

 

[Shannabel]

Yep!

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

 

[I like Serena]

Not so fast!

Let's start with:

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.

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?

 

[Shannabel]

Not so fast!

Let's start with:



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

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?

 

[I like Serena]

ohhh
i should have had
(du/u^2)+(du/u)
which integrates to
-1/u + ln u?

Hmm, let's take a step back.

You had:

then i get
du/4x^4(u) ... but i can't integrate that?

do you mean like x^4 = u+1?

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

 

[Shannabel]

Hmm, let's take a step back.

You had:



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

du/[4(u+1)(u)]?

 

[I like Serena]

du/[4(u+1)(u)]?

Yes!

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

 

[Shannabel]

Yes!

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?

 

[I like Serena]

i can take the 1/4 out right?

Yes!

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

No.

(Sorry for the repetition! )

 

[Shannabel]

Yes!



No.

(Sorry for the repetition! )

hmm.. okay then i think i am stumped :(

 

[I like Serena]

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)?

 

[Shannabel]

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)?

 

[I like Serena]

so (1/4) * du/u-du/(u+1)?

Yep!

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

 

[Shannabel]

Yep!

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

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
?

 

[Shannabel]

Yep!

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

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?

 

[I like Serena]

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

 

[I like Serena]

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?

Yep!

EDIT: err, no. The limits are not right.

 

[Shannabel]

Yep!

EDIT: err, no. The limits are not right.

what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81
= 1/4 [ln((-15/16)/(1/16))-ln((-80/81)/(1/81))]
= 1/4 [ln(-15)-ln(-80)]
=1/4 [ln(15/80)]
= 1/4ln(3/16)

:)

 

[Bohrok]

what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81

The limits 1/3 and 1/2 would've been wrong if you used them while the integral was in terms of u, but you changed them right here and those are the correct limits; you didn't tell us you changed them in the previous post

 

[I like Serena]

What Bohrok said!