# 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.