Express in partial fraction 2x+1/(x-1)(2x^ 3 +6x^2 - x+4)

Answer 1

Assuming the usual carelessness with parentheses,
(2x+1)/((x-1)(2x^3+6x^2-x+4))
This is going to be awkward, since the cubic has no rational roots. I guess we could go with
(2x+1)/((x-1)(2x^3+6x^2-x+4)) = A/(x-1) + (Bx^2+Cx+D)/(2x^3+6x^2-x+4)
That will give us
A(2x^3+6x^2-x+4) + (Bx^2+Cx+D)(x-1) = 2x+1
2Ax^3 + 6Ax^2 - Ax + 4A + Bx^3 - Bx^2 + Cx^2 -Cx + Dx - D = 2x+1
2A -B = 0
6A-B+C = 0
-A-C+D = 2
4A-D = 1
Solve all that to finish it off.