1. Make the substitution t = (u-3)^2 - 8
2. Try integration by parts, so that inside the integral sign, you differentiate ln(x) and integrate sqrt(x), the product of these two will be a power function.
3. Make the substitution s = u-3, multiply out the square, and then you will have a sum of power functions.
Here's an easier substitution in the first...
$\displaystyle \displaystyle \begin{align*} \int{\frac{1}{3 + \sqrt{t + 8}}\,dt} &= \int{\frac{2\sqrt{t + 8}}{2\sqrt{t + 8}\left( 3 + \sqrt{t+ 8}\right)} \,dt} \\ &= \int{\frac{2u}{3 + u}\,du} \textrm{ after making the substitution }u = \sqrt{t + 8} \implies du = \frac{1}{2\sqrt{t + 8}}\,dt \\ &= \int{\frac{2(v - 3)}{v}\,dv} \textrm{ after making the substitution } v = 3 + u \implies dv = du \\ &= \int{\frac{2v - 6}{v}\,dv} \\ &= \int{2 - \frac{6}{v}\,dv} \end{align*}$
Continue...