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

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

$=2intln(x)1/xdx$

$=2intudu$

$=2u^2/2+C$

$=u^2+C$

$=ln^2(x)+C$

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