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#