Evaluate the integral

Question

∫(3x)/(x²+6)dx

I used u-substitution for the first part, u=x²+6, du=2xdx, 1/2du=xdx

I put the substitutions back into the equation and took out the constant from the integral, getting 3/2∫1/udu (3/2 from 1/2du and the 3*x from the original equation, where xdx was substituted for 1/2du)

Except if I do that, I'd get 1/udu, which would be u^-1. But if I evaluate it, then it would be u⁰, right? So I'm not sure if I just got something wrong along the way or if I'm doing it completely wrong.

Answer 1

All looks fine.

You just need this fact: ∫1/u du = ln(u) +c