How do you integrate $\int \sqrt{x} ln 2 x d x$ $int sqrtx ln 2xdx$ ?

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Answer 1 · 2015-04-12T12:48:23

By part :

$d u = \sqrt{x}$ $du = sqrt(x)$
$u = \frac{2}{3} \sqrt{x^{3}}$ $u = 2/3sqrt(x^3)$

$v = ln (2 x)$ $v = ln(2x)$
$d v = \frac{1}{x}$ $dv = 1/x$

$= [\frac{2}{3} \sqrt{x^{3}} \cdot ln (2 x)] - \frac{2}{3} \int \frac{\sqrt{x^{3}}}{x} d x$ $=[2/3sqrt(x^3)*ln(2x)]-2/3intsqrt(x^3)/xdx$

Note : $\frac{\sqrt{x^{3}}}{x} = x^{\frac{3}{2} - 1} = x^{\frac{1}{2}} = \sqrt{x}$ $sqrt(x^3)/x = x^(3/2-1)=x^(1/2) = sqrt(x)$

Finally we have :

$= [\frac{2}{3} \sqrt{x^{3}} \cdot ln (2 x)] - \frac{4}{9} [\sqrt{x^{3}}] + C$ $=[2/3sqrt(x^3)*ln(2x)]-4/9[sqrt(x^3)]+C$

$= \frac{2}{9} \sqrt{x^{3}} (3 ln (2 x) - 2) + C$ $=2/9sqrt(x^3)(3ln(2x)-2)+C$

How do you integrate int sqrtx ln 2xdx∫√xln2xdxint sqrtx ln 2xdx?