 # 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 $\frac{1}{y\left(1-\frac{y}{L}\right)}$, cancel common terms in the numerator and denominator:
$=\int -\frac{L}{y\left(y-L\right)}dy$
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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It’s not a method, it simply do this:
$\int \frac{1}{y\left(1-\frac{y}{L}\right)}dx=\int \frac{1}{y\left(\frac{L-y}{L}\right)}dx=\int \frac{1}{y\frac{1}{L}\left(L-y\right)}dx=\int \frac{1}{\frac{1}{L}}\frac{1}{y\left(L-y\right)}dx\underset{\frac{1}{\frac{1}{L}}=L}{\underset{⏟}{=}}\int \frac{L}{y\left(L-y\right)}dx=\int \frac{L}{-y\left(y-L\right)}dx=\int -\frac{L}{y\left(y-L\right)}dx$

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It's algebra.
$\frac{1}{1-\frac{y}{L}}\cdot \frac{L}{L}=\frac{L}{L-y}=-\frac{L}{y-L}$
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 $\frac{1}{L}$; when you "cancel" that from $1$ you get $L$.

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