How do you integrate #int (lnx)^2# by parts?
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The answer is #=x((lnx)^2-2lnx+2)+C#
The integration by parts is
#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#
#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#
