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

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

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Answer 1 · 2018-06-27T10:49:00

$\int \frac{ln x}{x^{\frac{1}{2}}} d x = 2 \sqrt{x} (ln x - 2) + C$ $int lnx/x^(1/2) dx = 2 sqrtx(lnx - 2)+C$

Explanation:

Integrate by parts:

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

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

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

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

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

$\int \frac{ln x}{x^{\frac{1}{2}}} d x = 2 x^{\frac{1}{2}} ln x - 4 x^{+ \frac{1}{2}} + C$ $int lnx/x^(1/2) dx = 2 x^(1/2)lnx - 4 x^(+1/2)+C$

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

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

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Explanation:

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