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

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

2 Answers
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

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