Homework Help: Stuck at partial fraction problem.

1. Oct 2, 2011

Alfy102

1. The problem statement, all variables and given/known data
Evaluate ∫1/x(x-1)2, by using the partial fraction method

2. The attempt at a solution

Stating in partial fraction form:

1/(x(x-1)2 = A/x + B/(x-1) + C/(x-1)2

1 = Ax(x-1)2/x + Bx(x-1)2/(x-1) + Cx(x-1)2/(x-1)2

1 = A(x-1)2 + Bx(x-1) + Cx

and this is where I am stuck because I must compare the coefficients to 1...any idea how I can go on after this?

2. Oct 2, 2011

EkaterinaAvd

you have to express your expression like
1 = A(x-1)^2 + Bx(x-1) + Cx = Dx^2 + Ex + F

express D,E,F in terms of A,B and known numbers.

and you know that (Dx^2 + Ex + F = 1) for any x. To make it always equal to 1, your coefficients near x^2 and x must be 0 and your free coefficient must be 1.

3. Oct 2, 2011

epenguin

That is exactly what you do. You'd have to expand and collect first to know the coefficients of x0, x1, x2.

A bit pedestrian. As this is an identity true for all x, what happens when you make x = 1, when you make x = 0 ?

4. Oct 2, 2011

Alfy102

ok, continuin with the given information from the people above:

1 = A(x2-2x+1) + Bx2-B + Cx

1 = Ax2-2Ax+A + Bx2-B + Cx

Sorting through, then I will get:

1 = Ax2 + Bx2 -2Ax + Cx + A - B

1 = (A+B)x2 - (2A-C)x + (A-B)

From the above equation:

A + B = 0

2A - C = 0

A - B = 1

Therefore A=1/2 , B= -1/2, C = 1

Rewriting the partial fraction form:

∫1/x(x-1)2 dx = ∫ (1/2)/x + (-1/2)/(x-1) + 1/(x-1)2 dx

= ∫ x/2 dx + ∫ -(x-1)/2 dx + ∫ 1/(x-1)2 dx

= x2/4 - (x2/4 - x/2) + ∫1/(x-1)2

Do I have to re-apply partial fraction at the last part as well?

5. Oct 2, 2011

Kreizhn

You can certainly collect like terms: as a matter of fact that's how I used to do things also. However, epenguin described a really useful device for getting the coefficients in a much faster way.

Since you are trying to solve 1 = A(x-1)^2 + Bx(x-1) + Cx, this must hold for all x right? Well, try plugging x = 0 into this to find that (-1)^2A =1. Similarly, plugging in x=1 will give you C(1) =1. You can use these to solve for B.

Your way is still correct, but this way can save you the task of simplifying and avoid computational errors.

Edit: Just saw that the 2 was changed to a squared. Fixing my post to reflect that.

Last edited: Oct 2, 2011
6. Oct 2, 2011

Kreizhn

To which part? $\int\frac1{(x-1)^2} dx$? This can be solved with substitution.

Edit: Made a mistake with my comment

Last edited: Oct 2, 2011
7. Oct 2, 2011

Alfy102

All right, thanks guys. I've also done the substitution part.

u=x-1
du=dx

Finally got it.

At Kreizhn: Its "1 = A(x-1)^2 + Bx(x-1) + Cx", not "1 = A(x-1)2 + Bx(x-1) + Cx" or maybe you meant something else? :P

8. Oct 2, 2011

Kreizhn

Thanks for point that out. I actually copied and pasted that line, so it must have been changed from somewhere. Anyway, I've fixed that post to reflect that fact. Also note that, as I warned earlier, you made a computational mistake in evaluating your partial fractions. You expanded Bx(x-1) = Bx^2 - B which is not correct. Give the method I mentioned a try and you'll see the answers practically throw themselves at you.

9. Oct 2, 2011

Alfy102

oh dayum...dayum....d.a.y.u.m....

Thank Kreizhn for pointing that out, ouch.

So from the method that you were saying:

1= A(x-1)2 + Bx(x-1) + Cx

Substitute x=0 to the equation, I will get:

1 = A(0-1)2 + B(0)(0-1) + C(0)

1= A(-1)^2

A=1

Substitute x =1 to the equation, I will get:

1 = A(1-1)2 + B(1)(1-1) + C(1)

C=1

10. Oct 2, 2011

Kreizhn

Exactly. And you can use any other value to get an equation for B. Try x=2, that works nicely.

11. Oct 2, 2011

Alfy102

So, substituting x = 2, A = 1 and C = 1, I will get:

1 = A(2-1)^2 + B(2)(2-1) + C(2)

1 = A^2 + 2B + 2C

1= 1^1 + 2B + 2

2B = -2

B = -1

Got it, thx a lot. Much appreciated. Learned something new here. When you say "can use any other value to get an equation for B", can it be implemented to find A and B as well? or I must use x = 0 and x = 1 to initially find A and C?

12. Oct 2, 2011

Kreizhn

You can use anything. It's just that x=0 and x=1 were particularly useful because they happened to "kill" some of the other variables. In general, you won't be able to get rid all but one variable, but this gives a really easy way of getting the system of equations without needed to collect like terms.

13. Oct 2, 2011

Alfy102

Ok, thanks for the info.