Question

How do you find the integral of `int lnx/x^3 dx` from 1 to infinity?

Answer 1

`1/4`

Explanation:

`int udv = uv- int vdu`

`u=lnx => du=dx/x`

`dv=dx/x^3 => v= int dx/x^3=int x^-3dx= x^-2/-2=-1/(2x^2)`

`I=int_1^oo lnx/x^3 dx= -lnx/(2x^2) - int_1^oo (-1/(2x^2)) dx/x`

`I=-lnx/(2x^2)+1/2 int_1^oo dx/x^3 = [-lnx/(2x^2)+1/2(-1/(2x^2))]|_1^oo`

`I= [-lnx/(2x^2)-1/(4x^2)]|_1^oo`

`I=lim_(x->oo) (-lnx/(2x^2)-1/(4x^2)) - (-1/4)`

`I=1/4 - lim_(x->oo) lnx/(2x^2) -lim_(x->oo) 1/(4x^2)`

`I=1/4 - lim_(x->oo) (1/x)/(4x) - 0 = 1/4 - lim_(x->oo)1/(4x^2) = 1/4-0 = 1/4`