Integral by partial fractions?
- Thread starter summergrl
- Start date
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
I am willing to work this out for you, because I think that partial fractions are dinosaurs in calculus. Any CAS and most advanced calculators will give you the answer.
\(\displaystyle \L \frac{{x^3 - 4x - 10}}{{x^2 - x - 6}} = x + 1 + \frac{2}{{x + 2}} + \frac{1}{{x - 3}}\).
Why do we still ask students to do this? What a waste of time!
\(\displaystyle \L \frac{{x^3 - 4x - 10}}{{x^2 - x - 6}} = x + 1 + \frac{2}{{x + 2}} + \frac{1}{{x - 3}}\).
Why do we still ask students to do this? What a waste of time!
When you divide you get:
\(\displaystyle \L\\\frac{3x-4}{x^{2}-x-6}+x+1\)
You need the partial expansion for the first part.
Factor the denominator:
\(\displaystyle \L\\\frac{3x-4}{(x-3)(x+2)}\)
\(\displaystyle \L\\\frac{A}{x-3}+\frac{B}{x+2}=3x-4\)
\(\displaystyle \H\\A(x+2)+B(x-3)=3x-4\)
\(\displaystyle \L\\Ax+2A+Bx-3B=3x-4\)
Equate coefficients:
\(\displaystyle \L\\A+B=3\)
\(\displaystyle \L\\2A-3B=-4\)
A=1 and B=2
\(\displaystyle \L\\\frac{1}{x-3}+\frac{2}{x+2}\)
\(\displaystyle \L\\\frac{3x-4}{x^{2}-x-6}+x+1\)
You need the partial expansion for the first part.
Factor the denominator:
\(\displaystyle \L\\\frac{3x-4}{(x-3)(x+2)}\)
\(\displaystyle \L\\\frac{A}{x-3}+\frac{B}{x+2}=3x-4\)
\(\displaystyle \H\\A(x+2)+B(x-3)=3x-4\)
\(\displaystyle \L\\Ax+2A+Bx-3B=3x-4\)
Equate coefficients:
\(\displaystyle \L\\A+B=3\)
\(\displaystyle \L\\2A-3B=-4\)
A=1 and B=2
\(\displaystyle \L\\\frac{1}{x-3}+\frac{2}{x+2}\)
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
If you are lazy like me you can find the partial fraction without solving any equations. The trick is to expand the fraction around the singular points and adding up all the expansions in which you only include the singular terms. Liouville's theorem then guarantees that the resulting expression is the desired partial fraction decomposition
We have:
\(\displaystyle f(x)=\frac{x^3-4x-10}{x^2-x-6}=\frac{x^3-4x-10}{(x-3)(x+2)}\)
Expand around the point \(\displaystyle x=3\). Near \(\displaystyle x=3\) we have:
\(\displaystyle f(x)=\frac{A}{x-3}\) + higher order terms.
Multiply both sides by \(\displaystyle x-3\) and take the limit \(\displaystyle x\rightarrow 3\). The limit is trivial and you find A = 1. The expansion around \(\displaystyle x=-2\) is
\(\displaystyle f(x)=\frac{B}{x+2}\) + higher order terms.
And B is found to be 2 by multiplying both sides by \(\displaystyle x+2\) and taking the limit \(\displaystyle x\rightarrow -2\) (which is again trivial).
Finally we must expand around \(\displaystyle x=\infty\). This is only necessary if the degree of the numerator is equal or larger than that of the denominator. Regular terms in this expansion have negative powers of x, singular terms have positive powers of x. The result can be obtained using long division, just like Galactus did above. But now you don't need to consider the remainder. You find the leading terms of the expansion around \(\displaystyle x=\infty\) are
\(\displaystyle x+1\)
Adding up all three expansions reproduces \(\displaystyle f(x)\):
\(\displaystyle f(x)=x+1+\frac{1}{x-3}+\frac{2}{x+2}\)
We have:
\(\displaystyle f(x)=\frac{x^3-4x-10}{x^2-x-6}=\frac{x^3-4x-10}{(x-3)(x+2)}\)
Expand around the point \(\displaystyle x=3\). Near \(\displaystyle x=3\) we have:
\(\displaystyle f(x)=\frac{A}{x-3}\) + higher order terms.
Multiply both sides by \(\displaystyle x-3\) and take the limit \(\displaystyle x\rightarrow 3\). The limit is trivial and you find A = 1. The expansion around \(\displaystyle x=-2\) is
\(\displaystyle f(x)=\frac{B}{x+2}\) + higher order terms.
And B is found to be 2 by multiplying both sides by \(\displaystyle x+2\) and taking the limit \(\displaystyle x\rightarrow -2\) (which is again trivial).
Finally we must expand around \(\displaystyle x=\infty\). This is only necessary if the degree of the numerator is equal or larger than that of the denominator. Regular terms in this expansion have negative powers of x, singular terms have positive powers of x. The result can be obtained using long division, just like Galactus did above. But now you don't need to consider the remainder. You find the leading terms of the expansion around \(\displaystyle x=\infty\) are
\(\displaystyle x+1\)
Adding up all three expansions reproduces \(\displaystyle f(x)\):
\(\displaystyle f(x)=x+1+\frac{1}{x-3}+\frac{2}{x+2}\)
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
summergrl said:
You need to put absolute value signs around the arguments of the logarithms:
\(\displaystyle \L\int\frac{dx}{x-a}=\log\left |x-a\right |\)
This follows from the fact that the derivative of \(\displaystyle \log(-x)\) is \(\displaystyle \frac{1}{x}\). You only need to make sure that the integration path does not cross the singularity of the fraction. In this case the integration interval is from zero to 1 and the singularities are at minus 2 and at 3.