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#