How do you go from $\frac{1}{u(u+1)}$ to $\frac{1}{u}-\frac{1}{u+1}$?

Question

I have seen my teacher many times going from $\frac{1}{u(u+1)}$ to $\frac{1}{u}-\frac{1}{u+1}$.

How is that done? I mean if I reverse it I understand but how can I go from the first to the second if I don't know the second?

What methodology should I use?

Answer 1

(uppps: this has crossed with pedja's answer)

You begin with an assumtion. You assume, with some unknowns a and b

$\qquad \displaystyle{a \over u} + {b \over u+1 } = {1 \over u(u+1) } $

Then it must be that in the numerator of the product
$\quad \displaystyle a(u+1)+bu = 1 \to (a+b)u+a = 1 \text{ for all } u $

But for all u this can only be possible if $\small (a+b)=0$ and $\small a=1 \to b=-1$, thus you find the only solution for your assumtion

$\qquad \displaystyle{1 \over u} + {-1 \over u+1 } = {1 \over u(u+1) } $

Answer 2

$\frac{1}{u(u+1)}=\frac{a}{u}+\frac{b}{u+1} \Rightarrow 1=a(u+1)+bu \Rightarrow a+b=0 $ and $a=1$

Answer 3

$\frac{1}{u}-\frac{1}{u+1}=\frac{u+1}{u(u+1)}-\frac{u}{u(u+1)}=\frac{u+1-u}{u(u+1)}=\frac{1}{u(u+1)}$

the rest is experience. If you work a lot with formulas, you recognize some patterns, and then you can do the reverse as well.

Answer 4

The method is called "partial fraction decomposition".