How do you integrate #ln(x) x^ (3/2) dx#?

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Answer 1 · 2016-05-06T01:00:27

#intln(x)x^(3/2)dx=2/5x^(5/2)(ln(x)-2/5)+C#

For this problem, we will use integration by parts:

Let #u = ln(x)# and #dv = x^(3/2)dx#

Then #du = 1/xdx# and #v = 2/5x^(5/2)#

Applying the formula #intudv = uv - intvdu# gives us

#intln(x)x^(3/2)dx = 2/5x^(5/2)ln(x)-int2/5x^(5/2)*1/xdx#

#=2/5x^(5/2)ln(x)-2/5intx^(3/2)dx#

#=2/5x^(5/2)ln(x) - 2/5(2/5x^(5/2))+C#

#=2/5x^(5/2)(ln(x)-2/5)+C#