Answer
$$8\operatorname{arcsec} 4 - \frac{1}{2}\sqrt {15} - \frac{{2\pi }}{3} + \frac{1}{2}\sqrt 3 $$
Work Step by Step
$$\eqalign{
& \int_2^4 {x\operatorname{arcsec} x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = \operatorname{arcsec} x,{\text{ }}du = \frac{1}{{x\sqrt {{x^2} - 1} }}dx \cr
& dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& {\text{Integration by parts formula }} \cr
& \int {\underbrace {\operatorname{arcsec} x}_u} \underbrace {xdx}_{dv} = \underbrace {\left( {\operatorname{arcsec} x} \right)}_u\underbrace {\left( {\frac{{{x^2}}}{2}} \right)}_v - \int {\underbrace {\left( {\frac{{{x^2}}}{2}} \right)}_v} \underbrace {\left( {\frac{1}{{x\sqrt {{x^2} - 1} }}} \right)dx}_{du} \cr
& \int {x\operatorname{arcsec} x} dx = \frac{{{x^2}}}{2}\operatorname{arcsec} x - \frac{1}{2}\int {\frac{x}{{\sqrt {{x^2} - 1} }}} dx \cr
& \int {x\operatorname{arcsec} x} dx = \frac{{{x^2}}}{2}\operatorname{arcsec} x - \frac{1}{4}\int {\frac{{2x}}{{\sqrt {{x^2} - 1} }}} dx \cr
& \int {x\operatorname{arcsec} x} dx = \frac{{{x^2}}}{2}\operatorname{arcsec} x - \frac{1}{4}\left( {\frac{{\sqrt {{x^2} - 1} }}{{1/2}}} \right) + C \cr
& \int {x\operatorname{arcsec} x} dx = \frac{{{x^2}}}{2}\operatorname{arcsec} x - \frac{1}{2}\sqrt {{x^2} - 1} + C \cr
& {\text{Therefore,}} \cr
& \int_2^4 {x\operatorname{arcsec} x} dx = \left[ {\frac{{{x^2}}}{2}\operatorname{arcsec} x - \frac{1}{2}\sqrt {{x^2} - 1} } \right]_2^4 \cr
& = \left[ {\frac{{{4^2}}}{2}\operatorname{arcsec} 4 - \frac{1}{2}\sqrt {{4^2} - 1} } \right] - \left[ {\frac{{{2^2}}}{2}\operatorname{arcsec} 2 - \frac{1}{2}\sqrt {{2^2} - 1} } \right] \cr
& = \left[ {8\operatorname{arcsec} 4 - \frac{1}{2}\sqrt {15} } \right] - \left[ {2\left( {\frac{\pi }{3}} \right) - \frac{1}{2}\sqrt 3 } \right] \cr
& = 8\operatorname{arcsec} 4 - \frac{1}{2}\sqrt {15} - \frac{{2\pi }}{3} + \frac{1}{2}\sqrt 3 \cr} $$
