Question

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

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

Answer 1

`-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 2

`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`