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7dot4IntpolyOVERpoly

# 7dot4IntpolyOVERpoly - Supplement to 7.4 After talking to...

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Supplement to 7.4 After talking to many of you in office hours, I decided to write this to clarify some integration tricks that come up at the end of Partial Fractions problems. Oftentimes, the last step involves evaluating an integral of the form 2 ax b dx cx dx e + + + . I will work through a few examples to illustrate the principles involved. The two identities to keep in mind are ln du u C u = + and ( ) 1 1 2 2 tan x a a du C u a = + + Problem: 2 2 1 x dx x + + Solution: Split the fractions into two parts. ( ) ( ) 2 2 2 2 1 1 2 2 1 2 1 1 ln 1 2tan x dx x x dx dx x x x x C + = + + = + + + + + Make sure you see how to get each of the last two integrals. Problem: 2 2 2 2 x dx x x + + + Solution: Looking ahead to 2 2 2 w x x = + + , you would want the numerator to be 2 2 dw x = + . We massage the fraction into two parts; one having the correct numerator and the other being amenable to the 1 tan identity above. First, look at the x . We want that to be a 2 x , so we multiply the numerator by 2 and offset it with a fraction in the front.

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7dot4IntpolyOVERpoly - Supplement to 7.4 After talking to...

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