Partial fractions

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Given a function like

$\frac{-1}{x-1}+\frac{1}{x+1}$

we can rewrite it over a common denominator like this:

$\left ( \frac{-1}{x-1} \right )\left ( \frac{x+1}{x+1} \right )+\left ( \frac{1}{x+1} \right )\left ( \frac{x-1}{x-1} \right ) \\ \begin{align*} &=\frac{-x-1+x-1}{(x-1)(x+1)} \\ &= \frac{-2}{x^2-1} \end{align*}$

But note that the original form is easily integrated to give

$\int \left ( \frac{-1}{x-1}+\frac{1}{x+1} \right )dx \\ \begin{align*} &= -\textrm{In }(x-1)+ \textrm{In}(x+1)+c \end{align*}$

While faced with the form -2/(x^2-1) , we wouldn’t have known how to integrate it.

Note that the original function was of the form $(-1) / \cdots +(+1) /\cdots$ It’s not a coincidence that the two constants on top, −1 and +1, are opposite in sign but equal in absolute value. To see why, consider the behavior of this function for large values of x. Looking at the form -1/(x-1)+1/(x+1) , we might naively guess that for a large value of such as 1000, it would come out to be somewhere on the order thousandths. But looking at the form -2/(x^2-1) , we would expect it to be way down in the millionths. This seeming paradox is resolved by noting that for large values of , the two terms in the form -1/(x-1)+1/(x+1) very nearly cancel. This cancellation could only have happened if the constants on top were opposites like plus and minus one.

The idea of the method of partial fractions is that if we want to do an integral of the form $\int \frac{dx}{P(x)}$

where P(x) is an nth order polynomial, we rewrite 1/P as

$\frac{1}{P(x)}=\frac{A_1}{x-r_1}+\cdots \frac{A_n}{x-r_n}$

where r_1...r_n are the roots of the polynomial, i.e., the solutions of the equation P(r)=0 . If the polynomial is second-order, you can find the roots r_1 and r_2 using the quadratic formula; I'll assume for the time being that they're real. For higher-order polynomials, there is no surefire, easy way of finding the roots by hand, and you'd be smart simply to use computer software to do it. In Yacas, you canfind the real roots of a polynomial like this:

FindRealRoots(x^4-5*x^3
-25*x^2+65*x+84)
f3.,7.,-4.,-1.g

(I assume it uses Newton's method to nd them.) The constants A_i can then be determined by algebra, or by the following trick.

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Partial fractions