Consider $ \int f' \left( x \right) \operatorname{op} \left( g \left( x \right) \right) dx $ where f, f' and g are rational functions and op is either log, arc-trig or inverse hyperbolic function. Set u = f(x) and v = g(x). Then integrate by parts, resulting in $ \int \operatorname{op} \left( v \right) du = u \operatorname{op} \left( v \right) - \int u \operatorname{op}' \left( v \right) dv. $
Let op be ln, we obtain $ \int \ln v \, du = u \ln v - \int \frac u v dv. $
With this technique, we can even integrate $ \int \frac {\ln \left| x^2 + 2x \right|} {x^2 + 2x + 1} dx. $
For brevity, constants of integration are omitted. It is easy to get $ \int \left( x+3 \right) dx = \frac {x^2} 2 + 3x. $ Therefore, \begin{align} \int \left( x+3 \right) \ln x \, dx &= \left( \frac{x^2}2 + 3x \right) \ln x - \int \left( \frac x 2 + 3 \right) dx \\ &= \left( \frac{x^2}2 + 3x \right) \ln x - \frac {x^2 + 12x} 4. \end{align}
The rest is easy. $ \int_1^2 \left( x+3 \right) \ln x \, dx = 8 \ln 2 - \frac {15} 4. $