Find the partial fractions of (x^2+x-3)/((x-2)(x^2+2))

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Using partial fractions we can write the term as;

`(x^2+x-3)/((x-2)(x^2+2)) = A/(x-2)+(Bx+C)/(x^2+2)`

`(x^2+x-3)/((x-2)(x^2+2)) = (A(x^2+2)+(Bx+C)(x-2))/((x-2)(x^2+2))`

`x^2+x-3 = Ax^2+2A+(Bx+C)(x-2)`

`x^2+x-3 = Ax^2+2A+Bx^2-2Bx+Cx-2C`

`x^2+x-3 = (A+B)x^2+(C-2B)x+2A-2C`

Comparing components;

`x^2 rarr 1 = A+B ---(1)`

`x rarr 1 = C-2B ---(2)`

cons. `rarr -3 = 2A-2C ----(3)`

`(1)xx2+(2)`

`2+1 = 2A+C`

`3 = 2A+C ----(4)`

Solving (3) and (4) gives;

`C = 2`

`A = 1/2`

From (1) we can get `B = 1/2`

*So the answer is;*

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

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