Partial Fractions Integration Technique
Smaridasa: The integration technique we will learn now is the Partial Fractions Integration Technique. This technique is used when the integral involves rational functions (fractions) that cannot be easily integrated using other techniques like algebraic substitution, integration by parts, or trigonometric substitution. The main idea is to break down a complex fraction into simpler fractions that we know how to integrate.
Note: This technique is rarely discussed at the high school level but sometimes appears in university entrance exams. With proper practice and understanding, you can master it!
Basic Formulas
The partial fractions technique usually leads to integrals of the form:
1) \(\int \frac{k}{ax+b} dx = \frac{k}{a} \ln(ax+b) + C\)
2) \(\int \frac{k}{(ax+b)^n} dx = \frac{k}{a} \cdot \frac{1}{n+1} (ax+b)^{n+1} + C\)
Logarithm Properties
\(\ln a + \ln b = \ln(a \cdot b)\)
\(\ln a - \ln b = \ln\left(\frac{a}{b}\right)\)
How to Break Down Fractions (Partial Fractions)
The following are common patterns for breaking down fractions:
Case 1: Distinct linear factors
\(\frac{f(x)}{(a_1x+b_1)(a_2x+b_2)} = \frac{A}{a_1x+b_1} + \frac{B}{a_2x+b_2}\)
Case 2: Repeated linear factors
\(\frac{f(x)}{(a_1x+b_1)(a_2x+b_2)^2} = \frac{A}{a_1x+b_1} + \frac{B}{a_2x+b_2} + \frac{C}{(a_2x+b_2)^2}\)
Case 3: Repeated linear factors (higher power)
\(\frac{f(x)}{(a_1x+b_1)(a_2x+b_2)^3} = \frac{A}{a_1x+b_1} + \frac{B}{a_2x+b_2} + \frac{C}{(a_2x+b_2)^2} + \frac{D}{(a_2x+b_2)^3}\)
Case 4: Quadratic factor that can't be factored
\(\frac{f(x)}{(a_1x+b_1)(a_3x^2+b_3x+c_3)} = \frac{A}{a_1x+b_1} + \frac{Bx+C}{a_3x^2+b_3x+c_3}\)
Case 5: Repeated quadratic factor
\(\frac{f(x)}{(a_1x+b_1)(a_3x^2+b_3x+c_3)^2} = \frac{A}{a_1x+b_1} + \frac{Bx+C}{a_3x^2+b_3x+c_3} + \frac{Dx+E}{(a_3x^2+b_3x+c_3)^2}\)
Example Problems
1) Break down the following fractions into simpler forms:
a) \(\frac{2x+1}{x^2-3x}\)
b) \(\frac{x-3}{x^2-2x-8}\)
Smaridasa:
a) \(\frac{2x+1}{x^2-3x}\)
1) Factor the denominator: \(x^2-3x = x(x-3)\)
2) Break down: \(\frac{2x+1}{x(x-3)} = \frac{A}{x} + \frac{B}{x-3}\)
3) Combine: \(\frac{A}{x} + \frac{B}{x-3} = \frac{A(x-3)+Bx}{x(x-3)} = \frac{(A+B)x-3A}{x(x-3)}\)
4) Compare numerators: \(2x+1 = (A+B)x-3A\)
5) Solve system:
\(-3A = 1 \Rightarrow A = -\frac{1}{3}\)
\(A+B = 2 \Rightarrow -\frac{1}{3} + B = 2 \Rightarrow B = \frac{7}{2}\)
6) Final form: \(\frac{2x+1}{x^2-3x} = \frac{-\frac{1}{3}}{x} + \frac{\frac{7}{2}}{x-3} = \frac{1}{6}\left(\frac{-2}{x} + \frac{21}{x-3}\right)\)
b) \(\frac{x-3}{x^2-2x-8}\)
1) Factor the denominator: \(x^2-2x-8 = (x+2)(x-4)\)
2) Break down: \(\frac{x-3}{(x+2)(x-4)} = \frac{A}{x+2} + \frac{B}{x-4}\)
3) Combine: \(\frac{A}{x+2} + \frac{B}{x-4} = \frac{A(x-4)+B(x+2)}{(x+2)(x-4)} = \frac{(A+B)x-4A+2B}{(x+2)(x-4)}\)
4) Compare numerators: \(x-3 = (A+B)x-4A+2B\)
5) Solve system:
\(A+B = 1\)
\(-4A+2B = -3\)
6) Solving: \(A = \frac{5}{6}, B = \frac{1}{6}\)
7) Final form: \(\frac{x-3}{x^2-2x-8} = \frac{\frac{5}{6}}{x+2} + \frac{\frac{1}{6}}{x-4} = \frac{1}{6}\left(\frac{5}{x+2} + \frac{1}{x-4}\right)\)
2) Find the integrals:
a) \(\int \frac{2x+1}{x^2-3x} dx\)
b) \(\int \frac{x-3}{x^2-2x-8} dx\)
Smaridasa:
a) Using the result from problem 1(a):
\(\int \frac{2x+1}{x^2-3x} dx = \int \frac{1}{6}\left(\frac{-2}{x} + \frac{21}{x-3}\right) dx\)
\(= \frac{1}{6} \left( \int \frac{-2}{x} dx + \int \frac{21}{x-3} dx \right)\)
\(= \frac{1}{6} (-2 \ln|x| + 21 \ln|x-3|) + C\)
\(= \frac{1}{6} (\ln x^{-2} + \ln (x-3)^{21}) + C\)
\(= \frac{1}{6} \ln \left( \frac{(x-3)^{21}}{x^2} \right) + C\)
b) Using the result from problem 1(b):
\(\int \frac{x-3}{x^2-2x-8} dx = \int \frac{1}{6}\left(\frac{5}{x+2} + \frac{1}{x-4}\right) dx\)
\(= \frac{1}{6} \left( \int \frac{5}{x+2} dx + \int \frac{1}{x-4} dx \right)\)
\(= \frac{1}{6} (5 \ln|x+2| + \ln|x-4|) + C\)
\(= \frac{1}{6} (\ln (x+2)^5 + \ln (x-4)) + C\)
\(= \frac{1}{6} \ln \left( (x+2)^5 (x-4) \right) + C\)
3) Find \(\int \frac{x^2 + x + 3}{x^3 - x^2 + 4x - 4} dx\)
Smaridasa:
1) Factor the denominator: \(x^3 - x^2 + 4x - 4 = (x-1)(x^2+4)\)
2) Break down: \(\frac{x^2 + x + 3}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4}\)
3) Combine: \(\frac{A}{x-1} + \frac{Bx+C}{x^2+4} = \frac{A(x^2+4) + (Bx+C)(x-1)}{(x-1)(x^2+4)}\)
\(= \frac{(A+B)x^2 + (C-B)x + 4A-C}{(x-1)(x^2+4)}\)
4) Compare numerators: \(x^2 + x + 3 = (A+B)x^2 + (C-B)x + 4A-C\)
5) Solve system:
\(A+B = 1\)
\(C-B = 1\)
\(4A-C = 3\)
6) Solving: \(A = 1, B = 0, C = 1\)
7) So: \(\frac{x^2 + x + 3}{x^3 - x^2 + 4x - 4} = \frac{1}{x-1} + \frac{1}{x^2+4}\)
8) Integrate: \(\int \frac{1}{x-1} dx + \int \frac{1}{x^2+4} dx\)
\(= \ln|x-1| + \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C\)
4) Find \(\int \frac{3x^2 - x}{(x+1)(x-1)^2} dx\)
Smaridasa:
1) Break down: \(\frac{3x^2 - x}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\)
2) Combine: \(\frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} = \frac{A(x-1)^2 + B(x+1)(x-1) + C(x+1)}{(x+1)(x-1)^2}\)
\(= \frac{(A+B)x^2 + (C-2A)x + (A-B+C)}{(x+1)(x-1)^2}\)
3) Compare numerators: \(3x^2 - x = (A+B)x^2 + (C-2A)x + (A-B+C)\)
4) Solve system:
\(A+B = 3\)
\(C-2A = -1\)
\(A-B+C = 0\)
5) Solving: \(A = 1, B = 2, C = 1\)
6) So: \(\frac{3x^2 - x}{(x+1)(x-1)^2} = \frac{1}{x+1} + \frac{2}{x-1} + \frac{1}{(x-1)^2}\)
7) Integrate: \(\int \frac{1}{x+1} dx + \int \frac{2}{x-1} dx + \int \frac{1}{(x-1)^2} dx\)
\(= \ln|x+1| + 2\ln|x-1| + \int (x-1)^{-2} dx\)
\(= \ln|x+1| + \ln(x-1)^2 + \frac{1}{-1}(x-1)^{-1} + C\)
\(= \ln[(x+1)(x-1)^2] - \frac{1}{x-1} + C\)
Steps for Partial Fractions Integration
From the examples above, the partial fractions technique involves these steps:
- Factor the denominator completely
- Break down the fraction into simpler parts according to the partial fractions rules
- Determine the constants (A, B, C, etc.) by comparing coefficients
- Integrate each simpler fraction using basic integration formulas or other techniques if needed
Smaridasa: This technique is actually more suitable for university-level mathematics, but in the new curriculum (Kurikulum 2013), it appears in advanced mathematics. With diligent practice and good learning resources, you can definitely master it!