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multiply with conjugate 5-root3 such that the denominator gets rational
(5-root3)^2/((5+root3)(5-root3)) = (25-10root3+3)/(25-3)=(28-10root3)/22 = 14/11 - 5root3/11
x^3-6x^2+9x=x(x^2-6x+9)
observe binomial formula (x-3)^2=(x^2-6x+9)
Thus x(x-3)^2 and the function has two points meeting the x-axis.
First step: integrate int df/dx = x^2+3x+c (never forget the constant!)
Second step: substitute the point in order to get c
1 = (1)^2+3*1+c -> c = 1-1-3=-3
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