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Partial fractions

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I got stuck on this integrate
\(\displaystyle \int_0^{\infty}\frac{2x-4}{(x^2+1)(2x+1)}\)
and my progress
\(\displaystyle \int_0^{\infty} \frac{2x-4}{(x^2+1)(2x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{2x+1}\)
then I get these equation that I can't solve
and I get these equation..
\(\displaystyle 2a+c=0\) that is for \(\displaystyle x^2\)
\(\displaystyle 2b+a=2\) that is for \(\displaystyle x\)
\(\displaystyle b+c=-4\) that is for \(\displaystyle x^0\)
What have I done wrong?

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: partial fractions

The only thing I see wrong (besides omitting the differential from your original integral) is the line:

\(\displaystyle \int_0^{\infty} \frac{2x-4}{(x^2+1)(2x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{2x+1}\)
You should simply write:

\(\displaystyle \frac{2x-4}{(x^2+1)(2x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{2x+1}\)

You have correctly determined the resulting linear system of equations. Can you choose and use a method with which to solve it?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: partial fractions

The only thing I see wrong (besides omitting the differential from your original integral) is the line:



You should simply write:

\(\displaystyle \frac{2x-4}{(x^2+1)(2x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{2x+1}\)

You have correctly determined the resulting linear system of equations. Can you choose and use a method with which to solve it?
Thanks for pointing that!:) I have actually no clue how to solve it, I don't know what method I should use.

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: partial fractions

My choice would be elimination. Try subtracting the third equation from the first, and this will eliminate $c$, then combine this result with the second equation and you have a 2X2 system in $a$ and $b$. Can you state this system?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: partial fractions

My choice would be elimination. Try subtracting the third equation from the first, and this will eliminate $c$, then combine this result with the second equation and you have a 2X2 system in $a$ and $b$. Can you state this system?
I made it like a matrice and solved it :) Thanks for the help and sorry for not posting the progress but now I get \(\displaystyle b=0, c=-4, a=2\) and that works fine when I put those value in the equation!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: partial fractions

Any valid method you choose is fine. Your solution is correct and now integration is a breeze. (Rock)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: partial fractions

\(\displaystyle 2x-4 = (ax+b)(2x+1)+c(x^2+1)\)

Try the following method to find the constants

First Let $x = -{1 \over 2}$ then you can easily find $c $

Second Let $x =0$ you can find $b$ since you are given $c$

Finally find $a$ given $b$ and $c$ .
 

TheBigBadBen

Active member
May 12, 2013
84
Re: partial fractions

The only thing I see wrong (besides omitting the differential from your original integral) is the line:



You should simply write:

\(\displaystyle \frac{2x-4}{(x^2+1)(2x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{2x+1}\)

You have correctly determined the resulting linear system of equations. Can you choose and use a method with which to solve it?
Here's a fancy little trick (more commonly used in complex analysis; similar to Zaid's) to get through these sorts of problems. Rather than solving a system of equations, one can simply "plug in some numbers" to get to the answer.

Let's begin at the point where we know that

\(\displaystyle \frac{2x-4}{(x^2+1)(2x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{2x+1}\)

First of all, multiplying both sides by \(\displaystyle 2x+1\), we have

\(\displaystyle \frac{2x-4}{(x^2+1)} = (2x+1)\frac{ax+b}{x^2+1}+ c\)

Now, having multiplied by our choice of term in the denominator, we plug in a value of x that makes this term zero. \(\displaystyle 2x+1=0\) when \(\displaystyle x=-\frac{1}{2}\), so plug in that value. The first term on the right becomes zero after multiplying, leaving you with:

\(\displaystyle \frac{2(-\frac{1}{2})-4}{((-\frac{1}{2})^2+1)} = c\)

simply plug in to find the answer (c = 4).
We can do something similar with \(\displaystyle x^2+1\). First of all, multiply both sides to get

\(\displaystyle \frac{2x-4}{2x+1} = ax+b+ (x^2+1)\frac{c}{2x+1}\)

Now, we plug in a value of x that makes this term become zero. In this case, we note that \(\displaystyle x^2+1=0\) when \(\displaystyle x=\pm i\). Choosing \(\displaystyle x=i\) and noting that the second term on the right multiplies to zero, this becomes

\(\displaystyle \frac{2i-4}{2i+1} = b + a i\)

Which, after some complex-number algebra, gives you a real and imaginary part corresponding to b and a. That is, the above evaluates to \(\displaystyle 0 + 2i\), telling you that a = 2 and b = 0.

This method is particularly useful when you only want to find a particular term without solving for the rest. Note that this method does not work for irreducible terms in the denominator taken to powers greater than 1; that requires a more subtle approach.