How do you integrate #int (lnx)^2# by parts?

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Answer 1 · 2016-12-21T13:59:07

The answer is #=x((lnx)^2-2lnx+2)+C#

#intu'vdx=uv-intuv'dx#

Let #v=(lnx)^2#, #=>#, #v'=(2lnx)/x#

#u'=1#, #=>#, #u=x#

Therefore,

#int(lnx)^2dx=x(lnx)^2-2intlnxdx#

We do theintegration by parts a second time

Let #v=lnx#, #=>#, #v'=1/x#

#u'=1#, #=>#, #u=x#

#intlnxdx=xlnx-intx*1/x*dx#

#=xlnx-x#

Therefore,

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

#=x(lnx)^2-2xlnx+2x+C#

Answer 2 · 2016-12-21T14:02:49

#int (lnx)^2dx = x(ln^2x-2lnx+2)+C#

The formula for integration by parts states that:

#int u*dv = u*v -int v*du#

In this case we take #u(x) = (lnx)^2# and #v(x) = x#, so that:

#int (lnx)^2dx = x(lnx)^2-int 2xlnx(1/x)dx= x(lnx)^2-2int lnxdx#

We solve this last integral again by parts:

#int lnx = xlnx - int x*(1/x)dx = xlnx -int dx = xlnx -x+C#

Plugging this in the previous result:

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