# Integrals

• Mar 28th 2006, 06:05 AM
roseh
Integrals
I was told the best way to solve this problem dx/x^2-x with the limits b=3

The second problem is dx/x^2-4x+4 with the limits b=1 and a=-1.

I sincerely thank you for your help.
• Mar 28th 2006, 07:30 AM
topsquark
Quote:

Originally Posted by roseh
I was told the best way to solve this problem dx/x^2-x with the limits b=3

The second problem is dx/x^2-4x+4 with the limits b=1 and a=-1.

I sincerely thank you for your help.

I understand the first problem to be:
$\int_2^3 dx \, \frac{1}{x^2-x}$

So. You want to decompose the fraction into linear (or nonfactorable quadratic) parts.
$\frac{1}{x^2-x}=\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$
Where A and B are some constants.

There are various and sundry ways to solve for A and B. My favorite is simply to add the two fractions and compare the numerator with the numerator of the original expression. So:
$\frac{A}{x}+\frac{B}{x-1}=\frac{A(x-1)+B(x)}{x(x-1)}$

So we know that $A(x-1)+B(x)=1$.

By expanding the LHS and matching like coefficients we get that A+B=0 and -A=1. So A=-1 and B=1. Finally:

$\int_2^3 dx \, \frac{1}{x^2-x}=\int_2^3 dx \, \frac{-1}{x} + \int_2^3 dx \, \frac{1}{x-1}$

$=-ln(|x|) \right | _2^3 + ln(|x-1|) \right| _2^3$

$=(-ln3+ln2)+(ln2-ln1) = -ln3+2ln2-0=ln(2^2/3)$

$=ln(4/3)$

-Dan
• Mar 28th 2006, 07:59 AM
topsquark
Quote:

Originally Posted by roseh
I was told the best way to solve this problem dx/x^2-x with the limits b=3

The second problem is dx/x^2-4x+4 with the limits b=1 and a=-1.

I sincerely thank you for your help.

$\int_{-1}^1 dx \, \frac{1}{x^2-4x+4}=\int_{-1}^1 dx \, \frac{1}{(x-2)^2}$

The fraction decomposition in this case is not
$\frac{1}{(x-2)^2} \neq \frac{A}{x-2}+\frac{B}{x-2}$
You will quickly see by adding the fractions on the RHS that this does not work. What we need for repeated linear factors is:
$\frac{1}{(x-2)^2}=\frac{A}{x-2}+\frac{Bx+C}{(x-2)^2}$

Again, I will add the fractions on the RHS and equate the numerator with the original expression.
$A(x-2)+Bx+C=1$
So A+B=0 and -2A+C=1. There is no single set of numbers A,B, and C that will work. The simplest is A=B=0, C=1 which gives you back your original problem. (This implies that partial fractions is NOT the simplest way to do the problem, which you may have already noticed! ;) )

In the spirit of partial fractions I will use A=-1, B=1, C=-1.

$\int_{-1}^1 dx \, \frac{1}{x^2-4x+4}=\int_{-1}^1 dx \, \frac{1}{(x-2)^2}$

$=\int_{-1}^1 dx \, \frac{-1}{x-2} + \int_{-1}^1 dx \, \frac{x-1}{(x-2)^2}$
I will put y = x - 2 in the second integral, so x = y + 2 and dx = dy:

$=-ln(|x-2|) \right | _{-1}^1 + \int_{-3}^{-1} dy \, \frac{y+1}{y^2}$

$=(-ln1+ln3)+ \int_{-3}^{-1} dy \, \frac{1}{y} + \int_{-3}^{-1} dy \, \frac{1}{y^2}$

$=-0+ln3+ln(|y|) \right | _{-3}^{-1} - \frac{1}{y} \right | _{-3}^{-1}$

$= ln3+(ln1-ln3)-(\frac{1}{-1} - \frac{1}{-3})$

$=0-(-1+\frac{1}{3})=\frac{2}{3}$

Of course, this would have been a whole lot simpler just to set y = x - 2 in the original integral. :)

Note: I didn't mention it explicitly... Remember to check that the integral doesn't cross any "improper" points. (ie. Make sure the integrand doesn't go either to zero or infinity.)

-Dan
• Mar 28th 2006, 08:18 AM
topsquark
The only case of partial fractions I have not mentioned (because it didn't come up) is the case of non-factorable quadratic factors. By "non-factorable" I mean the quadratic has no real zeros. (You CAN use complex factors, but I, as a Physicist, usually don't.) So how would you decompose:
$\frac{x^2+1}{2x^3+4x^2+3x+1}$

I happen to know (because I built the problem!) that $2x^3+4x^2+3x+1=(x+1)(2x^2+2x+1)$. So we can break the original fraction into:

$\frac{x^2+1}{2x^3+4x^2+3x+1}=\frac{A}{x+1}+\frac{B x+C}{2x^2+2x+1}$

We then add the fractions on the RHS and compare numerators:

$A(2x^2+2x+1)+(Bx+C)(x+1)=x^2+1$
Note that the quadratic factor always has a linear term over it.

So we get that 2A+B=1, 2A+(B+C)=0, and A+C=1. I get A=2, B=-3, and C=-1.

$\frac{x^2+1}{2x^3+4x^2+3x+1}=\frac{2}{x+1}+\frac{-3x+-1}{2x^2+2x+1}$
as you can check by adding fractions.

In general for repeated factors, we always up the degree of the polynomial in the numerator for every factor we have, so for:
$\frac{1}{x^4}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D x^2+Ex+F}{x^3}+\frac{Gx^3+Hx^2+Ix+J}{x^4}$

So if you have something like:
$\frac{1}{x^2(x^2-3x+1)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{Dx+E}{x^ 2-3x+1}$

-Dan
• Mar 28th 2006, 08:55 AM
roseh
Thanks soooooooo much! The explanation/breakdown was terrific.
• Mar 28th 2006, 10:23 AM
topsquark
Quote:

Originally Posted by roseh
Thanks soooooooo much! The explanation/breakdown was terrific.

-Dan
• Mar 29th 2006, 04:18 AM
topsquark

Something was bugging me about a previous post so I looked it up.

My method for the repeated linear factors is not correct. (Well, technically it will work, but it isn't the simplest form!)

If you have:
$\frac{x}{(x-2)^2}$
The partial fraction expansion is:
$\frac{x}{(x-2)^2}= \frac{A}{x-2}+ \frac{B}{(x-2)^2}$

Previously I had told you the numerator of the second term would be Bx+C, which will work I suppose, since you can always just set B=0, but the new expansion I just gave you will work better. In other words, for repeated linear factors the numerators are just constants.

Sorry for the confusion!

-Dan