Answer
$${\text{The integral is divergent}}$$
Work Step by Step
$$\eqalign{
& \int_4^\infty {\ln \left( {5x} \right)} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_4^\infty {\ln \left( {5x} \right)} dx = \mathop {\lim }\limits_{b \to \infty } \int_4^b {\ln \left( {5x} \right)} dx \cr
& \cr
& {\text{integrating }}\int_{}^{} {\ln \left( {5x} \right)} dx{\text{ by parts}} \cr
& \,\,\,\,\,\,\,\,\,u = \ln \left( {5x} \right),\,\,\,\,\,\,du = \frac{1}{x}dx \cr
& \,\,\,\,\,\,\,\,\,dv = dx,\,\,\,\,\,\,v = x \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\ln \left( {5x} \right)dx} = x\ln \left( {5x} \right) - \int {x\left( {\frac{1}{x}} \right)} dx \cr
& \int {\ln \left( {5x} \right)dx} = x\ln \left( {5x} \right) - \int {dx} \cr
& \int {\ln \left( {5x} \right)dx} = x\ln \left( {5x} \right) - x + C \cr
& \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_4^b {\ln \left( {5x} \right)} dx = \mathop {\lim }\limits_{b \to \infty } \left( {x\ln \left( {5x} \right) - x} \right)_4^b \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\left( {{\text{see page 388}}} \right) \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\left( {b\ln \left( {5b} \right) - b} \right) - \left( {4\ln \left( {20} \right) - 4} \right)} \right] \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = \left( \infty \right) - \left( {4\ln \left( {20} \right) - 4} \right) \cr
& = \infty \cr
& then \cr
& {\text{The integral is divergent}} \cr} $$
