What is the antiderivative of #ln(2x)/x^(1/2)#?

1 Answer
Jun 22, 2016

Answer:

# 2 sqrt{x} ln(2x) -4 sqrt(x) + C#

Explanation:

IBP using #int u v' = uv - int u' v#

#u = ln(2x), u' = 1/x#
#v' = x^{-1/2}, v = 2x^{1/2}#

#\implies 2 sqrt{x} ln(2x) - int (2 sqrt(x))/x \ dx#

#= 2 sqrt{x} ln(2x) - int 2/ sqrt(x) \ dx#

#= 2 sqrt{x} ln(2x) - 2 * 2 sqrt(x) + C#

#= 2 sqrt{x} ln(2x) -4 sqrt(x) + C#