How do you find the integral of #f(x)=sqrt(x)*ln(x)# using integration by parts?

1 Answer
Gió
Nov 7, 2015

I found: #2/3x^(3/2)[ln(x)-2/3]+c#

Explanation:

I would first integrate #sqrt(x)# (written as #x^(1/2)#) and then derive #ln(x)# to get:
#intsqrt(x)ln(x)dx=(x^(1/2+1))/(1/2+1)ln(x)-int(x^(1/2+1))/(1/2+1)1/xdx=#
#=(x^(3/2))/(3/2)ln(x)-int2/3x^(3/2)/xdx=#
#=2/3x^(3/2)ln(x)-2/3intx^(3/2-1)dx=#
where I used the fact that #x^a/x^b=x^(a-b)#
#=2/3x^(3/2)ln(x)-2/3intx^(1/2)dx=#
#=2/3x^(3/2)ln(x)-2/3(x^(1/2+1))/(1/2+1)+c=#
#=2/3x^(3/2)ln(x)-2/3(x^(3/2))/(3/2)+c=#
#=2/3x^(3/2)ln(x)-4/9(x^(3/2))+c=#
#=2/3x^(3/2)[ln(x)-2/3]+c#

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