integration-partial fractions

[summergrl]

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

I wound up getting A=0, B+C+D=1, C-2D=-2, B+D=-1
Is that right because I cant't figure out what each letter equals

 

[skeeter]

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

when you multiply out the right side and combine like terms, you get ...

(A+C)x³ + (-A+B-2C+D)x² + (A+C-2D)x + (-A+B+D)

equating coefficients, you get ...

A+C = 0
-A+B-2C+D = 1
A+C-2D = -2
-A+B+D = -1

solving the above system ...

A = 1, C = -1, B = -1, D = 1

I know it's a pain, but you're just going to have to be more careful in your calculations.

 

[pka]

Did you know that there is available a Computer Algebra System at a minimal cost less than $40 that will decompose any fraction into partial fractions. I for one think that partial fractions will go the way of finding the square root of 13 with the cellulose-graphite (paper & pencil) method.

I ask, ‘what is the point of doing partial fractions’?
I know people who can use partial fractions to find a integral but have no idea what that integral means! Now That is pointless!

 

[Count Iblis]

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

Why not read-off B by substituting x = 1? And by substituting x = i you can read-off C and D. See, partial fractions is only laborious because students are not taught efficient ways to calculate it. This one is so simple that you can almost do it in your head.

 

[summergrl]

that is much easier!
well im not sure if i can do it with this one
Integration from 0 to 1
x^4+1/(x^4+5x^2+4)

i got A=0
B=-1/3
C=0
D=1/3

i know i messed up somewhere on this problem and im trying to figure out where, if it was before this point or after

 

[Count Iblis]

that is much easier!
well im not sure if i can do it with this one
Integration from 0 to 1
x^4+1/(x^4+5x^2+4)

i got A=0
B=-1/3
C=0
D=1/3

i know i messed up somewhere on this problem and im trying to figure out where, if it was before this point or after

The degree of the numerator is the same as the degree of the denominator. If the numerator is of equal or higher degree than the denominator then you have to do long division first. However, that's only necessary if you want to do it in the oficial tedious way (another good reason to learn other ways to tackle such problems)

This is how you could solve this problem (I actually teach it in a slightly different way to more advanced students). You first write down the general form of the partial fractions decomposition, but don't do long division (can be tedious and thus a potential source of errors). Instead, you look at how the fuinction behaves at infinity. In this case the limit to infinity exists and is 1. This means that the partial fraction decomposition must be of the form:

\(\displaystyle \frac{x^4+1}{(x^2+1)(x^2+4)} = 1 + \frac{A_{0}+A_{1}x}{x^2+1} + \frac{B_{0}+B_{1}x}{x^2+4}\)

You can now multiply both sides by the denominator and derive the equations for A and B, by equating coefficients of powers of x. But this is tedious. You can also insert the values of x for which the denominator is zero in the equation just like I did above. This is much simpler. Note that exactly at those points the equation is not a priori required to be valid. But you can, of course take the limit that x approaches the critical values, which leads to the conclusion that it must be valid at the critical points too.

Simpler still, you don't even need to multiply both sides by the entire denominator. Just multiplying by the individual factors in the denominator and taking the limit for x approaching a zero of that factor will yield the partial fraction for that factor.

So, let's multiply both sides by \(\displaystyle x^2 + 1\) and take the limit \(\displaystyle x\rightarrow i\). You then get:

\(\displaystyle \frac{i^4+1}{i^2+4}=A_{0} + i A_{1}\)

This gives

\(\displaystyle A_{0} = \frac{2}{3}\)


\(\displaystyle A_{1}=0\)

Now, let's multiply both sides by \(\displaystyle x^2 + 4\) and take the limit \(\displaystyle x\rightarrow 2i\). You get:

\(\displaystyle \frac{16 i^4+1}{4i^2+1}=B_{0} + i B_{1}\)

And we see that:

\(\displaystyle B_{0} = -\frac{17}{3}\)


\(\displaystyle B_{1}=0\)

Note that we could have used the fact that the fraction is actually just a function of

\(\displaystyle x^2\). If we put \(\displaystyle x^2 = y\), then in terms of y you only have linear factors in the denominator. From this you can immediately conclude that the coefficients \(\displaystyle A_{1}\) and \(\displaystyle B_{1}\) are zero.