Partial fractions
Tonight's idea
Adding fractions is routine: \(\dfrac{1}{x-1} - \dfrac{1}{x+1} = \dfrac{2}{x^2-1}\). Running that backwards, splitting one fraction into simple pieces, is partial fractions. It is what makes \(\int \frac{dx}{x^2 - 1}\) possible: you know how to integrate \(1/(x - 1)\), but not \(1/(x^2 - 1)\) directly.
Factor the denominator, \(x^2 - 1 = (x-1)(x+1)\), and look for constants \(A\), \(B\) with
\[\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}, \qquad\text{so}\qquad 1 = A(x+1) + B(x-1).\]
The second equation must hold for every \(x\), so choose convenient ones. \(x = 1\) kills the \(B\) term: \(1 = 2A\), \(A = \tfrac12\). \(x = -1\) kills \(A\): \(B = -\tfrac12\). (The cover-up shortcut: to find \(A\), cover \((x-1)\) in the original fraction and set \(x = 1\) in what remains.) Then
\(\displaystyle\int \frac{dx}{x^2-1} = \frac12\int\frac{dx}{x-1} - \frac12\int\frac{dx}{x+1} = \frac12 \ln\left|\frac{x-1}{x+1}\right| + C.\)
When the denominator is a product of distinct linear factors, each factor gets one constant, and there are always enough equations to find them, provided the numerator has lower degree than the denominator. If it does not, divide first. Repeated factors need one term for every power, and quadratic factors that do not split need numerators of the form \(Bx + C\).
The same split sums series. Since \(\dfrac{1}{n(n+1)} = \dfrac1n - \dfrac{1}{n+1}\), the sum of the first \(N\) terms collapses:
\[\Big(1 - \tfrac12\Big) + \Big(\tfrac12 - \tfrac13\Big) + \cdots + \Big(\tfrac1N - \tfrac{1}{N+1}\Big) = 1 - \frac{1}{N+1}.\]
Every middle term cancels with its neighbour, like a collapsing telescope. As \(N \to \infty\) the sum is exactly 1.
Watch one: the integral from 2 to 3
Find \(\int_2^3 \frac{dx}{x^2-1}\) using the split above, dropping the absolute values since \(x - 1 > 0\) on \([2, 3]\):
\[\frac12\Big[\ln(x-1) - \ln(x+1)\Big]_2^3 = \frac12\Big[(\ln 2 - \ln 4) - (\ln 1 - \ln 3)\Big] = \frac12 \ln\frac{3}{2} \approx 0.203.\]
Sanity check: on \([2, 3]\) the integrand runs from \(\tfrac13\) down to \(\tfrac18\), so the area over a width of 1 should be around 0.2. It is.
Your turn
- Split into partial fractions: (a) \(\dfrac{1}{x(x+2)}\), (b) \(\dfrac{3x+1}{(x-1)(x+2)}\).
- Evaluate \(\int_1^3 \frac{dx}{x(x+2)}\) exactly.
- Show that \(\dfrac{1}{(2n-1)(2n+1)} = \dfrac12\Big(\dfrac{1}{2n-1} - \dfrac{1}{2n+1}\Big)\). Find the sum of the first \(N\) terms, \(\dfrac{1}{1\cdot3} + \dfrac{1}{3\cdot5} + \cdots\), and the sum to infinity.
- In \(\int \frac{x^2}{x^2-1}\,dx\) the numerator does not have lower degree. Divide first, then find \(\int_2^3 \frac{x^2}{x^2-1}\,dx\).
Stretch: three factors
Find \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}\). Hint: each term is half the difference of two consecutive terms of tonight's series \(\frac{1}{n(n+1)}\). Then the sum telescopes.
Answers
- (a) \(\dfrac12\Big(\dfrac1x - \dfrac{1}{x+2}\Big)\). (b) Cover-up with \(x = 1\) gives \(A = 4/3\), with \(x = -2\) gives \(B = (-5)/(-3) = 5/3\). So \(\dfrac{4/3}{x-1} + \dfrac{5/3}{x+2}\).
- \(\dfrac12\Big[\ln x - \ln(x+2)\Big]_1^3 = \dfrac12\Big[(\ln 3 - \ln 5) - (0 - \ln 3)\Big] = \dfrac12 \ln\dfrac95 \approx 0.294\).
- \(\dfrac12\Big(\dfrac{1}{2n-1} - \dfrac{1}{2n+1}\Big) = \dfrac12 \cdot \dfrac{(2n+1) - (2n-1)}{(2n-1)(2n+1)}\), and the numerator is 2. The sum to \(N\) terms telescopes to \(\dfrac12\Big(1 - \dfrac{1}{2N+1}\Big) = \dfrac{N}{2N+1}\). As \(N \to \infty\) this tends to \(\dfrac12\).
- \(\dfrac{x^2}{x^2-1} = 1 + \dfrac{1}{x^2-1}\). So the integral is \(\Big[x\Big]_2^3 + \dfrac12\ln\dfrac32 = 1 + \dfrac12 \ln\dfrac32 \approx 1.203\): the example's answer plus the width of the interval.
Stretch. \(\dfrac12\Big(\dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)}\Big) = \dfrac12\cdot\dfrac{(n+2) - n}{n(n+1)(n+2)} = \dfrac{1}{n(n+1)(n+2)}\). Summing from 1 to \(N\), everything cancels except the first and last: \(\dfrac12\Big(\dfrac{1}{1\cdot2} - \dfrac{1}{(N+1)(N+2)}\Big)\). The sum to infinity is \(\dfrac12 \cdot \dfrac12 = \dfrac14\). The same move handles \(\dfrac{1}{n(n+1)(n+2)(n+3)}\), whose sum is \(\dfrac{1}{18}\).