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

## Expert Answers int (ln x)^2dx

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

So let

u = (lnx)^2

and

dv = dx

Then, differentiate u and integrate dv.

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

and

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

and

dv_2 = dx

Then, differentiate u and integrate dv.

du_2 = 1/x dx

and

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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