How do you integrate $\frac{ln \sqrt{x}}{x}$ $[ln sqrt x] / x$ ?

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How do you integrate $\frac{ln \sqrt{x}}{x}$ $[ln sqrt x] / x$ ?

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Answer 1 · 2018-04-12T02:28:19

The integral equals $\frac{1}{4} {ln}^{2} x + C$ $1/4ln^2x + C$

Explanation:

We can rewrite using logarithm laws.

$I = \int \frac{ln (x^{\frac{1}{2}})}{x} d x$ $I = int ln(x^(1/2))/xdx$

$I = \int \frac{ln x}{2 x} d x$ $I =int lnx/(2x) dx$

We now let $u = ln x$ $u = lnx$ . Then $d u = \frac{1}{x} d x$ $du = 1/xdx$ and then $d x = x d u$ $dx= xdu$ .

$I = \int \frac{u}{2 x} \cdot x d u$ $I = int u/(2x) * x du$

$I = \int \frac{1}{2} u d u$ $I = int 1/2u du$

$I = \frac{1}{2} (\frac{1}{2} u^{2}) + C$ $I = 1/2(1/2u^2) + C$

$I = \frac{1}{4} {ln}^{2} x + C$ $I = 1/4ln^2x + C$

Hopefully this helps!

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How do you integrate $\frac{ln \sqrt{x}}{x}$ $[ln sqrt x] / x$ ?

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