`int_4^infty 1/(x(ln x)^3)dx=`

Substitute `u=ln x` `=>` `du=1/x dx,` `u_l=ln 4,` `u_u=ln infty=infty` (`u_l` and `u_u` denote lower and upper bound respectively).

`int_(ln 4)^infty1/u^3 du=-1/(2u^2)|_(ln4)^infty=-1/2(lim_(u to infty)1/u^2-1/(ln 4)^2)=-1/2(0-1/(ln 4)^2)=`

`1/(2(ln 4)^2)approx0.260171`

**As we can see the integral converges and its value is**
`1/(2(ln 4)^2).`

The image below shows graph of the function and area under it representing the value of the integral. Looking at the image we can see that the graph approaches -axis (function converges to zero) "very fast". This suggests that the integral should converge to some finite number.

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