Integral of x^2+2 / x^3 + 6x + 3
u = x^3+6x + 3
du = 3x^2 + 6 = (1/2) du
(1/2) integral of du/u = (1/2) ln|u| + c
= (1/2)ln |x^3+6x+3| + C
ja oder nein?
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Where does 1/2 come from ?
u = x^3+6x + 3 --> ok
du/dx = 3x^2 + 6 = 3(x^2+2) and this is the numerator.
So the integral is now
oh right, stupid mistake
so it would simply be 1/3 in place of the 1/2 and the rest is correct?
It seems that... yes ^^
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