The correct option is
A xsecx−log|tan(π/4+x/2)|+c∫x secx tanx dx= ∫ x(secx(tanx))dx
=x∫(secx tanx)dx −∫1∫(secx tanx)dx ....... [Using integration parts]
=xsecx−∫secx dx
=xsecx−log|secx+tan x|+c
=xsecx−log|tan(π/4+x/2)|+c ..... [∵secx+tanx=tan(π/4+x/2)]
∴∫xsecxtanx dx=xsecx−log|tan(π/4+x/2)|+c