Integration by Partial Fractions
Types and worked examples (converted from the uploaded notes).
Type 1: Distinct linear factors
When the denominator is expressible as a product of distinct linear factors.
Example 1
Evaluate \( \displaystyle \int \frac{x-1}{(x+1)(x-2)} \, dx \).
Write
\( \dfrac{x-1}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2} \)
Then
\( x-1 = A(x-2) + B(x+1) \)
Putting \(x=2\): \(1 = 3B \Rightarrow B=\tfrac{1}{3}\).
Putting \(x=-1\): \(-2 = -3A \Rightarrow A=\tfrac{2}{3}\).
Thus
\( \dfrac{x-1}{(x+1)(x-2)} = \dfrac{2/3}{x+1} + \dfrac{1/3}{x-2} \)
Integrate:
\( \int \frac{x-1}{(x+1)(x-2)} dx = \frac{2}{3}\ln|x+1| + \frac{1}{3}\ln|x-2| + C \)
Example 2
Evaluate \( \displaystyle \int \frac{2x-1}{(x-1)(x+2)(x-3)} \, dx \).
Assume
\( \dfrac{2x-1}{(x-1)(x+2)(x-3)} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3} \)
Multiply through and compare or use substitution at roots.
Putting \(x=-2\): \(-5 = ( -3)(-5)B \Rightarrow B=-\tfrac{1}{3}\).
Putting \(x=1\): \(1 = (3)(-2)A \Rightarrow A=-\tfrac{1}{6}\).
Putting \(x=3\): \(5 = 2\cdot5\;C \Rightarrow C=\tfrac{1}{2}\).
So
\( \dfrac{2x-1}{(x-1)(x+2)(x-3)}
= -\tfrac{1}{6}\dfrac{1}{x-1} -\tfrac{1}{3}\dfrac{1}{x+2} + \tfrac{1}{2}\dfrac{1}{x-3}
\)
Integrate:
\( \int \dfrac{2x-1}{(x-1)(x+2)(x-3)} dx
= -\tfrac{1}{6}\ln|x-1| -\tfrac{1}{3}\ln|x+2| + \tfrac{1}{2}\ln|x-3| + C
\)
Type II: Repeated linear factors
When denominator contains repeating linear factors.
Example 3
Evaluate \( \displaystyle \int \frac{3x+1}{(x-2)^2(x+2)} \, dx \).
Assume decomposition of the form
\( \dfrac{3x+1}{(x-2)^2(x+2)} = \dfrac{A}{x-2} + \dfrac{B}{(x-2)^2} + \dfrac{C}{x+2} \)
Solving (substitute roots & compare coefficients) gives \(B=\tfrac{4}{7}\), \(C=-\tfrac{5}{16}\), and \(A=\tfrac{5}{16}\).
Thus integrate to get logarithms and simple rational terms. (Detailed algebra steps in original notes.)
Example 4
This example in the notes follows the same repeated-factor method — substitute roots and compare coefficients to find constants, then integrate term-by-term to obtain log terms and rational terms.
Type III: Irreducible quadratic factors
When denominator has irreducible quadratic factors, use linear numerators for those parts.
Example 5
Evaluate \( \displaystyle \int \frac{8}{(x+2)(x^2+4)} \, dx \).
Decompose as
\( \dfrac{8}{(x+2)(x^2+4)} = \dfrac{A}{x+2} + \dfrac{Bx+C}{x^2+4} \)
Solving constants yields \(A=1\), \(B=-1\), \(C=2\).
So
\( \dfrac{8}{(x+2)(x^2+4)} = \dfrac{1}{x+2} + \dfrac{-x+2}{x^2+4} \)
\end{pre>
Integration gives
\( \int \dfrac{8}{(x+2)(x^2+4)} dx
= \ln|x+2| - \tfrac{1}{2}\ln(x^2+4) + \arctan\left(\tfrac{x}{2}\right) + C
\)
