What is the integral `int ln x/x^2 dx`



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Posted on (Answer #1)

The integral `int ln x/x^2 dx` has to be determined.

Use integration by parts which gives `int u dv = u*v - int v*du`

In `int ln x/x^2 dx` , `u = ln x` , `dv = 1/x^2 dx`

`v = int 1/x^2 dx = -1/x`

`du = 1/x dx`

`int ln x/x^2 dx` = `ln x*(-1/x) - int (-1/x)*(1/x) dx`

=> `-ln x/x -1/x`

The integral `int ln x/x^2 dx = -ln x/x -1/x + C`


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