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Summary Worked Examples of Integration by Partial Fraction Decomposition, Calculus II for Physical Sciences, University of Toronto Scarborough(UTSC) $8.39   Add to cart
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Summary Worked Examples of Integration by Partial Fraction Decomposition, Calculus II for Physical Sciences, University of Toronto Scarborough(UTSC)
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  • Course
  • Calculus II For Physical Sciences (MATA36)
  • Institution
  • University Of Toronto (U Of T )

Explanation with examples, of integrating by partial fraction decomposition for Calculus II for Physical Sciences, University of Toronto Scarborough(UTSC)

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  • June 9, 2022
  • 4
  • 2021/2022
  • Summary

Subjects

  • math
  • calculus
  • calculus ii
  • integral calculus
  • mathematics
  • uoft
  • utsc
  • integration
  • physical sciences
  • environmental sciences
  • partial fraction decomposition
  • calculus ii for physical sciences

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  • University of Toronto (U of T )
  • Physical And Environmental Sciences
  • Calculus II For Physical Sciences (MATA36)
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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

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