Evaluating Integral with Partial Fractions: x^2-x/(x^2-1)^2

[soccergal13]

Evaluate the integral of x^2-x/(x^2-1)^2 from 0 to 1.

* I know that I have to use partial fractions in order to make the integral integratable.


My attempt at partial fractions:

A/(x-1) + (B/(x+1)) + (Cx+D/(x^2-1)^2)

Is this setup right? (Once I have it set up correctly, I know how to actually use the partial fractions in evaluating the integral). I wasn't sure because you can factor the bottom of the original function out to (x-1)(x-1)(x+1)(x+1), and thus A/(x-1) + B/(x-1) + C/(x+1) + D/(x+1) but I don't remember ever encountering a problem that needed to be divided into 4 partial fractions...

Thanks for any help!

 

[HallsofIvy]

No. (x²-1)² factors as (x-1)²(x+1). However, the numeator also can be factored as x(x-1) so, for all x except x= 1, you have x/[(x-1)(x+1)². Because of the (x-1) you will need both A/(x-1). Because of the (x+1)² you will need both B/(x+1) and C/(x+1)².
\frac{x^2- x}{(x^2-1)^2}= \frac{x}{(x-1)(x+1)2}= \frac{A}{x-1}+ \frac{B}{x-1}+ \frac{C}{(x-1)^2}

 

[Antineutron]

No. (x²-1)² factors as (x-1)²(x+1). However, the numeator also can be factored as x(x-1) so, for all x except x= 1, you have x/[(x-1)(x+1)². Because of the (x-1) you will need both A/(x-1). Because of the (x+1)² you will need both B/(x+1) and C/(x+1)².
\frac{x^2- x}{(x^2-1)^2}= \frac{x}{(x-1)(x+1)2}= \frac{A}{x-1}+ \frac{B}{x-1}+ \frac{C}{(x-1)^2}

You mean, \frac{x^2- x}{(x^2-1)^2}= \frac{x}{(x-1)(x+1)2}= \frac{A}{x-1}+ \frac{B}{x+1}+ \frac{C}{(x+1)^2}

 

[Astronuc]

Yeah - there was a problem with that LaTeX expression. It was supposed to have been fixed but for some reason the correction didn't take.

 

[soccergal13]

Thank you very much, that makes a lot more sense, but shouldn't it be Cx+D over the
(x^2+1)^2?

Also, when integrating each partial fraction from 0 to 1 (which is the second part of the problem), what do you do with the A/(1-x), because its going to be some number times the ln(1-x), which will equal negative infinty when you plug one in...?

 

[Antineutron]

Thank you very much, that makes a lot more sense, but shouldn't it be Cx+D over the
(x^2+1)^2?

Also, when integrating each partial fraction from 0 to 1 (which is the second part of the problem), what do you do with the A/(1-x), because its going to be some number times the ln(1-x), which will equal negative infinty when you plug one in...?

why don't you solve the problem without the limits first then plug in the limits and if you still see something wrong show us the where you have problems. make sure you use the ln properties to your advantage.

 

[soccergal13]

solving for A, B & C I found, in order, 1/4, -1/4 & 1/2
(by multiplying each partial fraction by the whole quantity (x+1)^2(x-1)) and then making equations with the A, B, & C's.)
****The question remains of-- did I need to have Cx +D over the (x^2+1)^2 because since the bottom is a quadratic, the top needs to be a linear function?
****Going with the 1/4, -1/4, and 1/2, I now have (.25)/(x-1) + (-.25)/(x+1)+ (.5)/(x+1)^2. Integrating gives you .25ln(x-1) - (.25ln(x+1)) + (.5(whatever the antiderrivative of 1/(x+1)^2 is)) (which I'm also confused about). Clearly, once I plug the limits in, .25ln(1-1)= .25ln(0) which = negative infinity. help!

 

[Antineutron]

you integrated the last term wrong, its not ln. its integral of (x+1)^-2dx

 

[soccergal13]

okay that i understand but that doesn't solve the problem of negative inifinty! lol. gah. sorry, I'm just getting really, really frustrated with this problem. and also if i needed the Cx+D on top... i do really appreciate your help :-), you have no idea!

 

[Antineutron]

and also if i needed the Cx+D on top...

no because underneath the (x+1) is not an irreducible quadratic factor

Here is one that is irreducible x^2+1 try to factor it

 

[Antineutron]

I don't know what to say... I agree, the limit of 1 messes it up. maybe the person that made up the problem made a mistake?

 

[Gib Z]

Well it diverges over the integral of integration, so really this is an improper integral. They probably wanted the Cauchy Principal value or were just incorrect.