#int_e^oo(1-lnx)/x^2 dx#=?

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Answer 1 · 2018-06-27T03:49:33

#-1/e#

#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 2 · 2018-06-27T03:58:27

#int_e^oo (1-lnx)/x^2 dx=-1/e#

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#

#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#