How do you evaluate the integral $int (lnx)^2dx$ ?

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Answer 1 · 2017-01-07T09:54:23

The answer is $=x(ln∣x∣)^2-2xln∣x∣+2x+C$

$intuv'dx=uv-intu'vdx$

Here, $int(lnx)^2dx$

Let, $u=(lnx)^2$ , $=>$ , $u'=(2lnx)/x$

$v'=1$ , $=>$ , $v=x$

So,

$int(lnx)^2dx=x(lnx)^2-intx*(2lnx)/xdx$

$=x(lnx)^2-2intlnxdx$

We do the the integration by parts a second time

$u=lnx$ , $=>$ , $u'=1/x$

$v'=1$ , $=>$ , $v=x$

so,

$intlnx=xlnx-intx*1/xdx$

$=xlnx-intdx=xlnx-x$

Putting it alltogether

$int(lnx)^2dx=x(lnx)^2-2(xlnx-x)+C$

$=x(ln∣x∣)^2-2xln∣x∣+2x+C$

How do you evaluate the integral int (lnx)^2dxint (lnx)^2dx?