How do you integrate $\int \sqrt{x} ln x$ $int sqrtxlnx$ by integration by parts method?

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How do you integrate $\int \sqrt{x} ln x$ $int sqrtxlnx$ by integration by parts method?

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Answer 1 · 2017-01-29T02:22:16

$\int \sqrt{x} ln (x) d x = \frac{2}{3} x^{\frac{3}{2}} (ln (x) - \frac{2}{3}) + C$ $intsqrt(x)ln(x)dx=2/3x^(3/2)(ln(x)-2/3)+C$

Explanation:

Using integration by parts:

Let $u = ln (x) \Rightarrow d u = \frac{1}{x} d x$ $u = ln(x) => du = 1/xdx$
and $d v = x^{\frac{1}{2}} d x \Rightarrow v = \frac{2}{3} x^{\frac{3}{2}}$ $dv = x^(1/2)dx => v = 2/3x^(3/2)$

Using the integration by parts formula $\int u d v = u v - \int v d u$ $intudv = uv-intvdu$

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

$= \frac{2}{3} x^{\frac{3}{2}} ln (x) - \int \frac{2}{3} x^{\frac{3}{2}} \cdot \frac{1}{x} d x$ $=2/3x^(3/2)ln(x)-int2/3x^(3/2)*1/xdx$

$= \frac{2}{3} x^{\frac{3}{2}} ln (x) - \frac{2}{3} \int x^{\frac{1}{2}} d x$ $=2/3x^(3/2)ln(x)-2/3intx^(1/2)dx$

$= \frac{2}{3} x^{\frac{3}{2}} ln (x) - \frac{2}{3} (\frac{2}{3} x^{\frac{3}{2}}) + C$ $=2/3x^(3/2)ln(x)-2/3(2/3x^(3/2))+C$

$= \frac{2}{3} x^{\frac{3}{2}} (ln (x) - \frac{2}{3}) + C$ $=2/3x^(3/2)(ln(x)-2/3)+C$

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How do you integrate $\int \sqrt{x} ln x$ $int sqrtxlnx$ by integration by parts method?

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