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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- Apr 14th 2008, 10:26 AMshepherdm1270check my substitution to find indefinate integral
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? - Apr 14th 2008, 10:29 AMMoo
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 $\displaystyle \int \frac{1}{3} \frac{du}{u}$ - Apr 14th 2008, 10:38 AMshepherdm1270
oh right, stupid mistake

so it would simply be 1/3 in place of the 1/2 and the rest is correct? - Apr 14th 2008, 11:17 AMMoo
It seems that... yes ^^