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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Hello, 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 $\displaystyle \int \frac{1}{3} \frac{du}{u}$
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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