integral of 2x/2+x-x^2dx?

2x * dx / (2 + x - x^2)
-2x * dx / (x^2 - x - 2)

x^2 - x - 2 = 0
x = (1 +/- sqrt(1 + 8)) / 2
x = (1 +/- 3) / 2
x = -2/2 , 4/2
x = -1 , 2
x + 1 , x - 2 = 0

-2x / ((x + 1) * (x - 2))

A/(x + 1) + B/(x - 2) = (-2x + 0) / ((x + 1) * (x - 2))

A * (x - 2) + B * (x + 1) = -2x + 0
Ax + Bx - 2A + B = -2x + 0

A + B = -2
-2A + B = 0

2 * (A + B) - 2A + B = 2 * (-2) + 0
2A - 2A + 2B + B = -4
3B = -4
B = -4/3

A + B = -2
A - 4/3 = -6/3
A = -2/3

(-2/3) * dx / (x + 1) - (4/3) * dx / (x - 2) =>
(-2/3) * (dx / (x + 1) + 2 * dx / (x - 2))

Integrate

(-2/3) * (ln|x + 1| + 2 * ln|x - 2|) + C
(-2/3) * ln|(x + 1) * (x - 2)^2| + C

2x/(2 + x - x²) = -2x/(x² - x - 2)
2x/(2 + x - x²) = -2x/(x + 1)(x - 2)

Let 2x/(2 + x - x²) = -2x/(x + 1)(x - 2) = [A/(x + 1)] + [B/(x - 2)]
Then A(x - 2) + B(x + 1) = -2x

Put x = -1 : -3A = 2, then A = -2/3
Put x = 2 : 3B = -4, then B = -4/3

∫[2x/(2 + x - x²)] dx
= ∫(-2/3) [1/(x + 1)] dx + ∫(-4/3) [1/(x - 2) dx + C
= -(2/3) ln|x + 1| - (4/3) ln|x - 2| + C
= -(2/3) ln|x + 1| - 2 ln|x - 2| + C
= -(2/3) ln[|x + 1|(x - 2)²] + C

If you mean ∫ 2x dx / [ 2 + x - x² ]
then you have to be clear with presentation.

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