How do you integrate #int sqrtx ln 2x dx # using integration by parts?

1 Answer
Mar 7, 2018

#I=2/9*x^(3/2)[3ln(2x)-2]+C#

Explanation:

#color(red)(I=intu*vdx=u*intvdx-int(u^'*intvdx)dx)#
#I=intx^(1/2)*ln(2x)dx,#
#Take,u=ln(2x)and v=x^(1/2)#
#rArru^'=1/(2x)*2=1/xand intvdx=(x^(3/2))/(3/2)=2/3*x^(3/2)+c#
#I=ln(2x)*2/3*x^(3/2)-int1/x*2/3*x^(3/2)dx#
#I=2/3*x^(3/2)ln(2x)-2/3intx^(1/2)dx#
#I=2/3*x^(3/2)ln(2x)-2/3*(x^(3/2))/(3/2)+C#
#I=2/3*x^(3/2)ln(2x)-4/9*x^(3/2)+C#
#I=2/9*x^(3/2)[3ln(2x)-2]+C#