How do you integrate #int sqrtxlnx# by integration by parts method?

1 Answer
Jan 29, 2017

#intsqrt(x)ln(x)dx=2/3x^(3/2)(ln(x)-2/3)+C#

Explanation:

Using integration by parts:

Let #u = ln(x) => du = 1/xdx#
and #dv = x^(1/2)dx => v = 2/3x^(3/2)#

Using the integration by parts formula #intudv = uv-intvdu#

#intsqrt(x)ln(x)dx = intln(x)x^(1/2)dx#

#=2/3x^(3/2)ln(x)-int2/3x^(3/2)*1/xdx#

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

#=2/3x^(3/2)ln(x)-2/3(2/3x^(3/2))+C#

#=2/3x^(3/2)(ln(x)-2/3)+C#