Class Notes (808,754)
Mathematics (2,710)
MAT136H1 (776)
all (232)
Lecture

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

1 Page
91 Views

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: ( ) ( )( ) ( ) ( ) ( ) (
More Less

Related notes for MAT136H1

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.