Answer
$$3$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\left( {1 + \ln x} \right){x^x}} dx \cr
& {\text{Write }}{x^x}{\text{ using the base }}e,{\text{ then }} \cr
& {x^x} = {e^{x\ln x}} \cr
& {\text{Let }}u = {x^x} \cr
& {\text{ }}u = {e^{x\ln x}} \cr
& {\text{ }}du = {e^{x\ln x}}\left( {x\left( {\frac{1}{x}} \right) + \ln x} \right)dx \cr
& {\text{ }}du = {e^{x\ln x}}\left( {1 + \ln x} \right)dx \cr
& {\text{The new limits of integration are:}} \cr
& x = 1 \Rightarrow u = {1^1} = 1 \cr
& x = 2 \Rightarrow u = {2^2} = 4 \cr
& {\text{Applying the substitution}} \cr
& \int_1^2 {\left( {1 + \ln x} \right){x^x}} dx = \int_1^4 {du} \cr
& {\text{Integrating}} \cr
& = \left[ u \right]_1^4 \cr
& = 4 - 1 \cr
& = 3 \cr} $$
