What is #int ln(x^2)/x#?

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Answer 1 · 2016-03-17T13:13:18

#intln(x^2)/xdx = ln^2(x)+C#

We will proceed using substitution.

Let #u = ln(x) => du = 1/xdx#

Then we have

#intln(x^2)/xdx = int(2ln(x))/xdx#

#=2intln(x)1/xdx#

#=2intudu#

#=2u^2/2+C#

#=u^2+C#

#=ln^2(x)+C#