Evaluate the indefinite integral.
∫(x + 3)/(x^2+6x) dx
This is an online hw prob. that covers sections integration by parts and substitution in indefinite integrals. it looks to me that it fits into the formula ∫udv=uv-∫vdu if you change the original equation to (x+3)*(x^2+6x)^-1
The Attempt at a Solution
Doing this would give you u=x+3, dv=(x^2+6x)^-1, du=1, v=(-log(x+6)-log(x))/6
the final answer becomes (x+3)((-log(x+6)-log(x))/6)-((-(x+6)log(x+6)+xlog(x)-2x-6)/6)
this however is not correct answer. Am I using the wrong method?