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What is the integral `int ln x/x^2 dx`
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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`
Posted by justaguide on July 1, 2013 at 1:46 PM (Answer #1)
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