# check my substitution to find indefinate integral

• Apr 14th 2008, 11:26 AM
shepherdm1270
check 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, 11:29 AM
Moo
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 $\int \frac{1}{3} \frac{du}{u}$
• Apr 14th 2008, 11:38 AM
shepherdm1270
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, 12:17 PM
Moo
It seems that... yes ^^