7.4 Integration of Rational Functions by Partial Fractions Question #3 (Medium)
University of Toronto St. George
7.4 Integration Techniques
Partial Fractions for Rational Functions
Question #3 (Medium): Partial Fractions of Irreducible Functions
Difference of cubes follows after the form of ( )( ). Notice the second term
is irreducible quadratic function.
Evaluate the integral using partial fractions.
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:
( ) ( )( ) ( ) ( ) ( ) (