Answer
$$4\ln 4 - 3$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\int_x^{{x^2}} {\frac{1}{y}} dydx} \cr
& = \int_1^4 {\left[ {\int_x^{{x^2}} {\frac{1}{y}} dy} \right]dx} \cr
& {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr
& = \int_x^{{x^2}} {\frac{1}{y}} dy \cr
& {\text{integrate}} \cr
& = \left[ {\ln \left| y \right|} \right]_x^{{x^2}} \cr
& {\text{evaluating the limits}} \cr
& = \ln \left| {{x^2}} \right| - \ln \left| x \right| \cr
& {\text{simplifying by using logarithmic properties}} \cr
& = \ln \left( {\frac{{{x^2}}}{x}} \right) \cr
& = \ln x \cr
& \cr
& \int_1^4 {\left[ {\int_x^{{x^2}} {\frac{1}{y}} dy} \right]dx} = \int_1^4 {\ln xdx} \cr
& {\text{integrating by parts }}\int u dv = uv - \int {vdu} \cr
& \,\,\,\,\,u = \ln x,\,\,\,du = \frac{1}{x}dx,\,\,\,\,\,dv = dx,\,\,\,\,\,\,v = x \cr
& \,\,\,\,\int {\ln x} dx = x\ln x - \int {x\left( {\frac{1}{x}} \right)} dx \cr
& \,\,\,\,\,\,\, = x\ln x - \int {dx} \cr
& \,\,\,\,\, = x\ln x - x + C \cr
& {\text{then}} \cr
& \int_1^4 {\ln xdx} = \left( {x\ln x - x} \right)_1^4 \cr
& {\text{evaluate}} \cr
& = \left( {4\ln 4 - 4} \right) - \left( {\ln 1 - 1} \right) \cr
& {\text{simplifying}} \cr
& = 4\ln 4 - 4 + 1 \cr
& = 4\ln 4 - 3 \cr} $$
