Question

what is the shortcut method for partial fractions ?

Answer 1

I'm not familiar with any names for any shortcuts, but I know of two methods for determining partial fractions.

Take \(\dfrac{1}{x(x+1)}\) as an example:
\[\begin{align*}\frac{1}{x(x+1)}&=\frac{A}{x}+\frac{B}{x+1}\\
1&=A(x+1)+Bx\end{align*}\]
Method 1: Expand any expandable terms, and group those with equal powers of the variable:
\[\begin{align*}1&=A(x+1)+Bx&&(1)\\
1&=Ax+A+Bx\\
1&=(A+B)x+A\end{align*}\]
Match up the coefficients, and you get a system of equations:
\[\begin{cases}A+B=0\\A=1\end{cases}~~\Rightarrow~~A=1,~B=-1\]
Method 2: Plug in values of \(x\) that will make some of the terms in \((1)\) disappear.

Let \(x=-1\), then
\[\begin{align*}1&=A(-1+1)+B(-1)\\
1&=0A-B\\
B&=-1\end{align*}\]
Let \(x=0\), then
\[\begin{align*}1&=A(0+1)+B(0)\\
1&=A+0B\\
A&=1\end{align*}\]
I think Method 2 is the shortcut. I prefer Method 1 because you can't always easily apply Method 2.

Answer 2

I also learnt about the "cover up method", but only works for certain cases. You can check it here http://math.mit.edu/suppnotes/suppnotes03/h.pdf

Answer 3

thank you. the cover up method is what im looking for!