What is the antiderivative of ${(ln x)}^{2}$ $(ln x)^2$ ?

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Answer 1 · 2015-02-19T22:45:26

Hello,

The answer is $x {(ln x)}^{2} - 2 x ln x + 2 x$ $x (ln x)^2 - 2x ln x + 2x$ .

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

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

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