University of Toronto Scarborough
MATA36
Calculus II for Physical Sciences
Text Book: Calculus Early Transcendentals 9th Edition by James
Stewart, Daniel Clegg, Saleem Watson
Partial fraction decomposition is the breaking down of a fraction into multiple fractions where
the numerator is a constant. The multiple fractions, which are obtained by decomposing, if we
add them, we get the original fraction. This helps in integrating the original fraction.
1
∫ ( x −3)( x+5) dx
First, we decompose the given fraction.
The first step is to factorize the denominator, here, the denominator is already factorized.
1
( x−3)(x +5)
A B
+
( x−3) ( x+ 5)
A and B are two constants, the value of which we have to find out. To do that, multiply both
sides of the equation (decomposed and undecomposed) by the factorized denominator.
1 A B
( x−3)( x +5)=( + )( x−3)( x +5)
( x−3 ) ( x+5 ) ( x−3 ) ( x+ 5 )
1= A ( x+ 5)+ B( x−3)
Now, one strategy is to input certain values to x which will remove one of the constants. For
example, if we input, x=3, then B (3-3) = 0, leaving only the constant A. Repeat the procedure
for the other constant.
Input x = 3,
1= A ( 3+5 )+ B ( 3−3 )
1=8 A
1
A=
8
MATA36
Calculus II for Physical Sciences
Text Book: Calculus Early Transcendentals 9th Edition by James
Stewart, Daniel Clegg, Saleem Watson
Partial fraction decomposition is the breaking down of a fraction into multiple fractions where
the numerator is a constant. The multiple fractions, which are obtained by decomposing, if we
add them, we get the original fraction. This helps in integrating the original fraction.
1
∫ ( x −3)( x+5) dx
First, we decompose the given fraction.
The first step is to factorize the denominator, here, the denominator is already factorized.
1
( x−3)(x +5)
A B
+
( x−3) ( x+ 5)
A and B are two constants, the value of which we have to find out. To do that, multiply both
sides of the equation (decomposed and undecomposed) by the factorized denominator.
1 A B
( x−3)( x +5)=( + )( x−3)( x +5)
( x−3 ) ( x+5 ) ( x−3 ) ( x+ 5 )
1= A ( x+ 5)+ B( x−3)
Now, one strategy is to input certain values to x which will remove one of the constants. For
example, if we input, x=3, then B (3-3) = 0, leaving only the constant A. Repeat the procedure
for the other constant.
Input x = 3,
1= A ( 3+5 )+ B ( 3−3 )
1=8 A
1
A=
8