Question

how do you do long division on this? x^2/(x^2-1)

Answer 1

but i need an answer that looks like 1+1/2(x-1)-1/2(x+1)

Answer 2

dafudge i think its partial LOL?

Answer 3

It's x^2/x^2-1 nt x/x^2-1 @.Sam

Answer 4

\[\huge \frac{x^2}{x ^{2}-1}\]
\[\huge \frac{x^2}{(x+1)(x-1)}\]
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\[\huge \frac{x^2}{(x+1)(x-1)}=\frac{A}{x+1}+\frac{B}{(x-1)}\]
\[\huge x^2=A(x-1)+B(x+1)\]
When x=1
B=1/2
When x=-1
A=-1/2
\[\huge \frac{x^2}{(x+1)(x-1)}=-\frac{1}{2(x+1)}+\frac{1}{2(x-1)}\]

Answer 5

\[\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}=\frac{x^2-1}{x^2-1}+\frac{1}{x^2-1}\]

\[=1+\frac{1}{x^2-1}=1+\frac{1}{(x+1)(x-1)}\]

then break up \(\displaystyle\frac{1}{(x+1)(x-1)}\)