What is #int xln(x)^2#?

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Answer 1 · 2016-03-20T13:01:45

Supposing you mean #ln(x)^2=(lnx)^2#

You have to integrate by parts twice. Answer is:

#x^2/2(ln(x)^2-lnx+1/2)+c#

Supposing you mean #ln(x)^2=ln(x^2)#

You have to integrate by parts once. Answer is:

#x^2(lnx-1/2)+c#

Supposing you mean #ln(x)^2=(lnx)^2#

#intxln(x)^2dx=#

#=int(x^2/2)'ln(x)^2dx=#

#=x^2/2ln(x)^2-intx^2/2(ln(x)^2)'dx=#

#=x^2/2ln(x)^2-intx^cancel(2)/cancel(2)*cancel(2)lnx*1/cancel(x)dx=#

#=x^2/2ln(x)^2-intxlnxdx=#

#=x^2/2ln(x)^2-int(x^2/2)'lnxdx=#

#=x^2/2ln(x)^2-(x^2/2lnx-intx^2/2(lnx)'dx)=#

#=x^2/2ln(x)^2-(x^2/2lnx-intx^cancel(2)/2*1/cancel(x)dx)=#

#=x^2/2ln(x)^2-(x^2/2lnx-1/2intxdx)=#

#=x^2/2ln(x)^2-(x^2/2lnx-1/2x^2/2)+c=#

#=x^2/2ln(x)^2-(x^2/2lnx-x^2/4)+c=#

#=x^2/2ln(x)^2-x^2/2lnx+x^2/4+c=#

#=x^2/2(ln(x)^2-lnx+1/2)+c#

Supposing you mean #ln(x)^2=ln(x^2)#

#intxln(x)^2dx=intx*2lnxdx#

#2intxlnxdx=#

#=2int(x^2/2)'lnxdx=#

#=2(x^2/2lnx-intx^2/2*(lnx)'dx)=#

#=2(x^2/2lnx-intx^cancel(2)/2*1/cancel(x)dx)=#

#=2(x^2/2lnx-1/2intxdx)=#

#=2(x^2/2lnx-1/2x^2/2)+c=#

#=cancel(2)*x^2/(cancel(2))(lnx-1/2)+c=#

#=x^2(lnx-1/2)+c#