decompose expression into a sum of partial fractions

6/`x^(2)` -9

partial fraction decomposition

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`6/(x^2-9) = 6/((x-3)(x+3))`

`6/((x-3)(x+3)) = A/(x-3)+B/(x+3)`

`6 = A(x+3)+B(x-3)`

`6 = (A+B)x+3(A-B)`

By omparing components;

`x rarr` `0 = A+B` `---(1)`

const. `rarr` `6 = 3(A-B) ----(2)`

By solving (1) and (2) we can get;

`A = 1`

`B = -1`

*So the partial fractions will give you;*

`6/(x^2-9) = 1/(x-3)-1/(x+3)`

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