How do you integrate $[ln sqrt x] / x$ ?

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Answer 1 · 2018-04-12T02:28:19

The integral equals $1/4ln^2x + C$

We can rewrite using logarithm laws.

$I = int ln(x^(1/2))/xdx$

$I =int lnx/(2x) dx$

We now let $u = lnx$ . Then $du = 1/xdx$ and then $dx= xdu$ .

$I = int u/(2x) * x du$

$I = int 1/2u du$

$I = 1/2(1/2u^2) + C$

$I = 1/4ln^2x + C$

Hopefully this helps!

How do you integrate [ln sqrt x] / x [ln sqrt x] / x?