What is the antiderivative of $\frac{ln (2 x)}{x^{\frac{1}{2}}}$ $ln(2x)/x^(1/2)$ ?

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What is the antiderivative of $\frac{ln (2 x)}{x^{\frac{1}{2}}}$ $ln(2x)/x^(1/2)$ ?

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Answer 1 · 2017-02-09T20:55:56

$= 2 x^{\frac{1}{2}} (ln (2 x) - 2) + C$ $= 2 x^(1/2) ( ln(2x) - 2) + C$

Explanation:

$\int \frac{ln (2 x)}{x^{\frac{1}{2}}} d x$ $int ln(2x)/x^(1/2) dx$

We set it up for IBP:
$= \int ln (2 x) \frac{d}{d x} (2 x^{\frac{1}{2}}) d x$ $= int ln(2x) d/dx(2 x^(1/2) ) dx$

Applying the IBP:
$= 2 x^{\frac{1}{2}} ln (2 x) - \int \frac{d}{d x} (ln (2 x)) (2 x^{\frac{1}{2}}) d x$ $= 2 x^(1/2) ln(2x) - int d/dx( ln(2x) )(2 x^(1/2) ) dx$

$= 2 x^{\frac{1}{2}} ln (2 x) - \int \frac{1}{x} \cdot 2 x^{\frac{1}{2}} d x$ $= 2 x^(1/2) ln(2x) - int 1/x * 2 x^(1/2) dx$

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

$= 2 x^{\frac{1}{2}} ln (2 x) - 2 \cdot 2 x^{\frac{1}{2}} + C$ $= 2 x^(1/2) ln(2x) - 2* 2 x^(1/2) + C$

$= 2 x^{\frac{1}{2}} (ln (2 x) - 2) + C$ $= 2 x^(1/2) ( ln(2x) - 2) + C$

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What is the antiderivative of $\frac{ln (2 x)}{x^{\frac{1}{2}}}$ $ln(2x)/x^(1/2)$ ?

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What is the antiderivative of $\frac{ln (2 x)}{x^{\frac{1}{2}}}$ $ln(2x)/x^(1/2)$ ?