Using long division, I get (x+1)(2x^2 + 7x - 6)

Now I can do something with that....

(x+3)/(2x^3+9x^2+13x+6) = A/(x+1) + (Bx+C)/(2x^2 + 7x - 6)

Multiply thru by the denominator on both sides....

2Ax^2 + 7Ax - 6A + (Bx+C)(x+1 ) = x + 3
2Ax^2 = 7Ax - 6A + Bx^2 + Bx + Cx + C = x + 3

Now we equate like terms.... So,

2Ax^2 + Bx^2 = 0 -> 2A + B = 0

Bx + Cx + 7Ax = x -> B + C + 7A = 1

And, -6A + C = 3

So, B = -2A and C = 3 + 6A

So, (-2A) + (3 + 6A) + 7A = 1 -> 11A = -2 -> A = -2/11

So, B = -2A = -2(-2/11) = 4/11

And C = 3 + 6A = 3 + 6(-2/11) = 21/11

So, (x+3)/(2x^3+9x^2+13x+6) = (-2/11)/(x+1) + ((4/11)x + (21/11))/(2x^2 + 7x - 6)

Hope that all makes sense to ya....

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