How do you integrate $int x^(1/2)*ln(x)$ using integration by parts?

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Answer 1 · 2015-11-29T03:56:03

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

Integration by part :
$intf(x)g'(x)dx = f(x)g(x)-intf'(x)g(x)dx$

Given $int root()x*ln(x)dx$

Let $f(x)= lnx$
$color(blue)(f'(x)= 1/x dx$ $" " " " " "(1)$

Let $g'(x)= x^(1/2)dx hArr g(x) = int x^(1/2)dx =>color(red)( g(x)= 2/3 x^(3/2)$ $" " " " " " (2)$

$int root()x*ln(x)dx$ = $color(red)(2/3 x^(3/2))*ln(x)- intcolor(red)( " " 2/3 x^(3/2)*color(blue)((1/x)dx)$

$2/3 x^(3/2) ln(x) - 2/3 intx^(1/2)dx$

$2/3 x^(3/2) ln(x) - 2/3*2/3(x^(3/2)) + C$

$2/3 x^(3/2) ln(x) - 4/9(x^(3/2)) + C$

Can be simplify to
$2/9 x^[3/2)(3ln(x)-2) +C$

How do you integrate int x^(1/2)*ln(x) int x^(1/2)*ln(x) using integration by parts?