Integration by Partial Fractions Technique
Smaridasa: This technique is also known as Partial Fractions. We use this technique for integration problems that are difficult to solve directly with other integration techniques such as "algebraic substitution", "integration by parts", and "trigonometric substitution". As the name suggests, we simplify fractional forms, especially the denominators.
Basic Formulas Used
The partial fractions technique usually leads to integrals of the form:
i) \[ \int k(ax + b)^n \, dx = \frac{k}{a} \cdot \frac{1}{n+1} (ax + b)^{n+1} + c \]
ii) \[ \int \frac{k}{ax + b} \, dx = \frac{k}{a} \ln(ax + b) + c \]
Properties of ln (read as "len") are the same as logarithm properties:
\[ \ln a + \ln b = \ln(a \cdot b) \]
\[ \ln a - \ln b = \ln\left(\frac{a}{b}\right) \]
How to Perform Partial Fractions
The following are common ways to break down fractions in problems:
Assume we have forms \((a_1x + b_1)\) and \((a_2x + b_2)\) that can no longer be factored, then the partial fraction decomposition is:
\[ \frac{f(x)}{(a_1x + b_1)(a_2x + b_2)} = \frac{A}{a_1x + b_1} + \frac{B}{a_2x + b_2} \]
\[ \frac{f(x)}{(a_1x + b_1)(a_2x + b_2)^2} = \frac{A}{a_1x + b_1} + \frac{B}{a_2x + b_2} + \frac{C}{(a_2x + b_2)^2} \]
\[ \frac{f(x)}{(a_1x + b_1)(a_2x + b_2)^3} = \frac{A}{a_1x + b_1} + \frac{B}{a_2x + b_2} + \frac{C}{(a_2x + b_2)^2} + \frac{D}{(a_2x + b_2)^3} \]
\[ \frac{f(x)}{(a_1x + b_1)(a_3x^2 + b_3x + c_3)} = \frac{A}{a_1x + b_1} + \frac{Bx + C}{a_3x^2 + b_3x + c_3} \]
\[ \frac{f(x)}{(a_1x + b_1)(a_3x^2 + b_3x + c_3)^2} = \frac{A}{a_1x + b_1} + \frac{Bx + C}{a_3x^2 + b_3x + c_3} + \frac{Dx + E}{(a_3x^2 + b_3x + c_3)^2} \]
We determine the values of A, B, C, D, E, etc. by equating the left and right sides.
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)
*) First, factor the denominator:
\[ x^2 - 3x = x(x - 3) \]
*) Break the fraction into two parts:
\[ \frac{2x+1}{x^2-3x} = \frac{2x+1}{x(x-3)} = \frac{A}{x} + \frac{B}{x-3} \]
\[ = \frac{A(x-3) + Bx}{x(x-3)} = \frac{(A+B)x - 3A}{x(x-3)} \]
Equate numerators:
\[ 2x + 1 = (A+B)x - 3A \]
Equate coefficients:
\[ -3A = 1 \rightarrow A = -\frac{1}{3} \]
\[ A + B = 2 \rightarrow -\frac{1}{3} + B = 2 \rightarrow B = \frac{7}{2} \]
Thus:
\[ \frac{2x+1}{x^2-3x} = \frac{-\frac{1}{3}}{x} + \frac{\frac{7}{2}}{x-3} = \frac{1}{6} \left( \frac{-2}{x} + \frac{21}{x-3} \right) \]
b)
*) Factor the denominator:
\[ x^2 - 2x - 8 = (x+2)(x-4) \]
*) Break the fraction:
\[ \frac{x-3}{x^2-2x-8} = \frac{x-3}{(x+2)(x-4)} = \frac{A}{x+2} + \frac{B}{x-4} \]
\[ = \frac{A(x-4) + B(x+2)}{(x+2)(x-4)} = \frac{(A+B)x - 4A + 2B}{(x+2)(x-4)} \]
Equate numerators:
\[ x - 3 = (A+B)x - 4A + 2B \]
Equate coefficients:
\[ A + B = 1 \quad \text{(i)} \]
\[ -4A + 2B = -3 \quad \text{(ii)} \]
Solve: Multiply (i) by 2: \( 2A + 2B = 2 \)
Subtract from (ii): \( -6A = -5 \rightarrow A = \frac{5}{6} \)
From (i): \( \frac{5}{6} + B = 1 \rightarrow B = \frac{1}{6} \)
Thus:
\[ \frac{x-3}{x^2-2x-8} = \frac{\frac{5}{6}}{x+2} + \frac{\frac{1}{6}}{x-4} = \frac{1}{6} \left( \frac{5}{x+2} + \frac{1}{x-4} \right) \]
2) Determine the integral results:
a) \(\int \frac{2x+1}{x^2-3x} \, dx\)
b) \(\int \frac{x-3}{x^2-2x-8} \, dx\)
Smaridasa:
a)
\[ \int \frac{2x+1}{x^2-3x} \, dx = \int \frac{1}{6} \left( \frac{-2}{x} + \frac{21}{x-3} \right) dx \]
\[ = \frac{1}{6} \left( -2 \int \frac{1}{x} dx + 21 \int \frac{1}{x-3} dx \right) \]
\[ = \frac{1}{6} \left( -2 \ln|x| + 21 \ln|x-3| \right) + c \]
\[ = \frac{1}{6} \ln \left( \frac{(x-3)^{21}}{x^2} \right) + c \]
b)
\[ \int \frac{x-3}{x^2-2x-8} \, dx = \int \frac{1}{6} \left( \frac{5}{x+2} + \frac{1}{x-4} \right) dx \]
\[ = \frac{1}{6} \left( 5 \int \frac{1}{x+2} dx + \int \frac{1}{x-4} dx \right) \]
\[ = \frac{1}{6} \left( 5 \ln|x+2| + \ln|x-4| \right) + c \]
\[ = \frac{1}{6} \ln \left[ (x+2)^5 (x-4) \right] + c \]
3) Determine the integral result: \(\int \frac{x^2 + x + 3}{x^3 - x^2 + 4x - 4} \, dx\)
Smaridasa:
*) Factor the denominator:
\[ x^3 - x^2 + 4x - 4 = (x-1)(x^2 + 4) \]
*) Partial fraction decomposition:
\[ \frac{x^2 + x + 3}{(x-1)(x^2 + 4)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 4} \]
\[ = \frac{A(x^2 + 4) + (Bx + C)(x-1)}{(x-1)(x^2 + 4)} \]
\[ = \frac{(A+B)x^2 + (C - B)x + 4A - C}{(x-1)(x^2 + 4)} \]
Equate numerators:
\[ x^2 + x + 3 = (A+B)x^2 + (C - B)x + 4A - C \]
Equate coefficients:
\[ A + B = 1 \quad \text{(i)} \]
\[ C - B = 1 \quad \text{(ii)} \]
\[ 4A - C = 3 \quad \text{(iii)} \]
Solve: From (i): \( B = 1 - A \)
From (iii): \( C = 4A - 3 \)
Substitute into (ii): \( (4A - 3) - (1 - A) = 1 \rightarrow 5A - 4 = 1 \rightarrow A = 1 \)
Then: \( B = 0 \), \( C = 1 \)
Thus:
\[ \frac{x^2 + x + 3}{x^3 - x^2 + 4x - 4} = \frac{1}{x-1} + \frac{1}{x^2 + 4} \]
*) Integrate:
\[ \int \frac{x^2 + x + 3}{x^3 - x^2 + 4x - 4} \, dx = \int \frac{1}{x-1} \, dx + \int \frac{1}{x^2 + 4} \, dx \]
\[ = \ln|x-1| + \frac{1}{2} \arctan\left(\frac{x}{2}\right) + c \]
4) Determine the integral result: \(\int \frac{3x^2 - x}{(x+1)(x-1)^2} \, dx\)
Smaridasa:
*) Partial fraction decomposition:
\[ \frac{3x^2 - x}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \]
\[ = \frac{A(x-1)^2 + B(x+1)(x-1) + C(x+1)}{(x+1)(x-1)^2} \]
\[ = \frac{(A+B)x^2 + (C - 2A)x + (A - B + C)}{(x+1)(x-1)^2} \]
Equate numerators:
\[ 3x^2 - x = (A+B)x^2 + (C - 2A)x + (A - B + C) \]
Equate coefficients:
\[ A + B = 3 \quad \text{(i)} \]
\[ C - 2A = -1 \quad \text{(ii)} \]
\[ A - B + C = 0 \quad \text{(iii)} \]
Solve: From (i): \( B = 3 - A \)
From (ii): \( C = 2A - 1 \)
Substitute into (iii): \( A - (3 - A) + (2A - 1) = 0 \rightarrow 4A - 4 = 0 \rightarrow A = 1 \)
Then: \( B = 2 \), \( C = 1 \)
Thus:
\[ \frac{3x^2 - x}{(x+1)(x-1)^2} = \frac{1}{x+1} + \frac{2}{x-1} + \frac{1}{(x-1)^2} \]
*) Integrate:
\[ \int \frac{3x^2 - x}{(x+1)(x-1)^2} \, dx = \int \frac{1}{x+1} dx + \int \frac{2}{x-1} dx + \int \frac{1}{(x-1)^2} dx \]
\[ = \ln|x+1| + 2 \ln|x-1| + \frac{(x-1)^{-1}}{-1} + c \]
\[ = \ln\left[(x+1)(x-1)^2\right] - \frac{1}{x-1} + c \]
Conclusion: The partial fractions technique is performed in steps: first factor the denominator, second decompose the fraction into parts according to the rules, third determine the constants by equating coefficients, and fourth integrate as needed (sometimes using basic formulas, sometimes requiring other integration techniques).
Smaridasa: This specialized integration technique is more common in college-level mathematics, but in the new curriculum (2013 Curriculum) it appears in advanced high school mathematics. With diligent study, enthusiasm, and good reference materials, students can master it well.