Unknown process in integration For the integrand 1/(y(1-y/L)), cancel common terms in the numerator and denominator: =int -L/(y(y-L))dy. This is part of WolframAlpha’s calculations before creating partial fractions. What exactly is this method/theorem? And how does it work?

tsuyakas1 2022-09-16 Answered
Unknown process in integration
For the integrand 1 y ( 1 y L ) , cancel common terms in the numerator and denominator:
= L y ( y L ) d y
This is part of WolframAlpha’s calculations before creating partial fractions. What exactly is this method/theorem? And how does it work?
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Answers (2)

faliryr
Answered 2022-09-17 Author has 15 answers
It’s not a method, it simply do this:
1 y ( 1 y L ) d x = 1 y ( L y L ) d x = 1 y 1 L ( L y ) d x = 1 1 L 1 y ( L y ) d x = 1 1 L = L L y ( L y ) d x = L y ( y L ) d x = L y ( y L ) d x

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moidu13x8
Answered 2022-09-18 Author has 2 answers
It's algebra.
1 1 y L L L = L L y = L y L
I suppose you could call it "multiplying by one"; it's a pretty common trick when working with fractions.
Calling it "canceling the common terms" is a little misleading, though. In this case, the common term is 1 L ; when you "cancel" that from 1 you get L.

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