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Lecture

# 7.4 Integration of Rational Functions by Partial Fractions Question #3 (Medium)

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School
University of Toronto St. George
Department
Mathematics
Course
MAT136H1
Professor
all
Semester
Winter

Description
7.4 Integration Techniques Partial Fractions for Rational Functions Question #3 (Medium): Partial Fractions of Irreducible Functions Strategy Difference of cubes follows after the form of ( )( ). Notice the second term is irreducible quadratic function. Sample Question Evaluate the integral using partial fractions. ∫ Solution The denominator is difference of cubes. First factor: ( )( ). The first term is linear, and the second quadratic but irreducible because since its discriminant . Decompose into partial fractions with coefficient: ( ) ( ). Then expand ( ) ( )( ) the numerator: . Here instead of expanding, it is better to substitute which makes ( )( ) so ( ) , so that . Then: ( ) ( )( ) ( ) ( ) ( ) (
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