∫ (x^2)/(2+x^3) dx

I have in my own work:
∫ x^2(2+x^3)^-1 dx
with u=2+x^3
du=3x^2
dx=du/3x^2
So ∫ x^2 (u)^-1 du/(3x^2)
∫ 1/3 (u)^-1
And that's where I get lost b/c the anti derivative of (u)^-1 is (u)^0 which can't be right. Please show your steps.

Question

∫ (x^2)/(2+x^3) dx

I have in my own work:
∫ x^2(2+x^3)^-1 dx
with u=2+x^3 
du=3x^2
dx=du/3x^2
So ∫ x^2 (u)^-1 du/(3x^2)
∫ 1/3 (u)^-1
And that's where I get lost b/c the anti derivative of (u)^-1 is (u)^0 which can't be right. Please show your steps.

lalaly · Answer

your right
but the last step$$\int\limits{\frac{1}{3u}du}=\frac{1}{3} \int\limits{\frac{du}{u}}=\frac{1}{3}\ln(u)$$

lalaly · Answer

because integral of 1/x =ln(x) not x^0

hartnn · Answer

lastly don't forget to re-substitute u= 2+x^3 in 1/3 ln u+c

anonymous · Answer

So my anti derivative of (u)^-1 would just be u/1? or ln(u)?

hartnn · Answer

antiderivative of u^-1 is ln u+c
not u/1 or u^0