What is the antiderivative of #(ln x)^2#?

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Answer 1 · 2015-02-19T22:45:26

Hello,

The answer is #x (ln x)^2 - 2x ln x + 2x#.

You have to know that antiderivative of #ln x# is #xln x - x + c#.

Use integration by parts : #int u'v = uv - int uv'#.

#int ln x \cdot ln x \ dx = ln x \cdot (x ln x - x) - int 1/ x (x ln x - x)#

So,

#int ln x \cdot ln x \ dx = x (ln x)^2 - x ln x - int (ln x - 1)#

and then,

#int ln x \cdot ln x \ dx = x (ln x)^2 - x ln x - (x ln x - x - x) + c#

and then #int ln x \cdot ln x \ dx = x (ln x)^2 - 2x ln x + 2x + c#