`int (ln x)^2 dx` Evaluate the integral

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`int (ln x)^2dx`

To evaluate, apply integration by parts `int udv= uv - vdu` .

So let

`u = (lnx)^2`


`dv = dx`

Then, differentiate u and integrate dv.

`du = 2lnx * 1/x dx = (2lnx)/x dx`


`v = intdx = x`

Plug-in them to the formula. So the integral becomes:

`int (ln x)^2dx`

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

`= x(lnx)^2 -2int lnx dx`

To take integral of ln x, apply integration by parts again.

So let 

`u_2 = ln x `


`dv_2 = dx`

Then, differentiate u and integrate dv.

`du_2 = 1/x dx`


`v_2= int dx = x`

So the integral becomes:

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

`= x(lnx)^2 - 2(xlnx - int dx)`

`=x(lnx)^2 - 2(xlnx - x)`

`=x(lnx)^2 - 2xlnx + 2x`


Therefore, `int (lnx)^2dx = x(lnx)^2 - 2xlnx + 2x` .

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