$\displaystyle

\int_1^3 (x+3)/(x^2+6x) dx = ?

$

I tired separating and simplifying the problem into two different integrals:

$\displaystyle

\int_1^3 1/(x+6) $ +$\displaystyle

\int_1^3 3/(x^2+6x)

$

and get...

$\displaystyle \int_1^3 ln(x+6) + \int_1^3 3ln(x^2+6x) $

evaluating I get

$\displaystyle

[ln9-ln7]+ 3[ln27+ln7]

$

but that is not any of the solutions.