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#