How do you integrate $int sqrtx ln 2xdx$ ?

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Answer 1 · 2015-04-12T12:48:23

By part :

$du = sqrt(x)$
$u = 2/3sqrt(x^3)$

$v = ln(2x)$
$dv = 1/x$

$=[2/3sqrt(x^3)*ln(2x)]-2/3intsqrt(x^3)/xdx$

Note : $sqrt(x^3)/x = x^(3/2-1)=x^(1/2) = sqrt(x)$

Finally we have :

$=[2/3sqrt(x^3)*ln(2x)]-4/9[sqrt(x^3)]+C$

$=2/9sqrt(x^3)(3ln(2x)-2)+C$

How do you integrate int sqrtx ln 2xdxint sqrtx ln 2xdx?