#int_e^oo(1-lnx)/x^2 dx#=? #int_e^oo(1-lnx)/x^2 dx#=? Calculus 2 Answers RedRobin9688 Jun 27, 2018 #-1/e# Explanation: #lim_(R->oo)int_e^R(1-lnx)/x^2dx# Let #u=1-lnx# and #dv=1/x^2dx# #du=-1/x# and #v=-1/x# #lim_(R->oo)int_e^R(1-lnx)/x^2dx=lim_(R->oo)(lnx-1)/x [x:R->oo]-int_e^R1/x^2dx# #lim_(R->oo)((lnx-1)/x [x:e->R]+1/x [x:e->R])# #lim_(R->oo) [((lnR-1)/R)-((lne-1)/e)]+[1/R-1/e]# #lim_(R->oo)[(color(blue)(d/dx)(lnR-1)/R)-0]+[0-1/e]# #lim_(R->oo)((1/R)/1)-1/e# #0-1/e# #-1/e# Answer link maganbhai P. Jun 27, 2018 #int_e^oo (1-lnx)/x^2 dx=-1/e# Explanation: Here, #I=int_e^oo (1-lnx)/x^2 dxto[becauselne=1]# #=int_e^oo(lne-lnx)/x^2 dxtoApply[lnA-lnB=ln(A/B)]# #=int_e^oo ln(e/x)/x^2dx# Subst . #e/x=u=>-e/x^2dx=du=>1/x^2dx=-1/edu# #xtoe=>uto1 and xto oo=>u to0# So, #I=-int_1^0 lnu*1/edu# #=1/eint_0^1 1*lnudu# Using Integration by parts: #I=1/e{[lnu*u]_0^1-int_0^1 1/u*udu}# #=1/e{[0-0]-int_0^1 1du}# #=1/e{-[u]_0^1}# #=-1/e[1-0]# #=-1/e# Answer link Related questions How do I determine the molecular shape of a molecule? What is the lewis structure for co2? What is the lewis structure for hcn? How is vsepr used to classify molecules? What are the units used for the ideal gas law? How does Charle's law relate to breathing? What is the ideal gas law constant? How do you calculate the ideal gas law constant? How do you find density in the ideal gas law? Does ideal gas law apply to liquids? Impact of this question 1415 views around the world You can reuse this answer Creative Commons License