How do you integrate #int xln x^3 dx # using integration by parts?

1 Answer
Dec 12, 2016

#I=(3x^2)/4[2lnx-1]+C#

Explanation:

need to use integration by parts.

formula is:#" "intu(dv)/(dx)dx=uv-intv(du)/(dx)dx#

the choice of #" "u" "#&#" "v# is vital.

if logs are involved then it is usual to take #u=lnf(x)# but try and simplify the expression using the laws of logs first if possible.

#intxlnx^3dx=int3xlnxdx#

#u=lnx=>(du)/(dx)=1/x#

#(dv)/(dx)=3x=>v=(3x^2)/2#

#I=(3x^2)/2lnx-int(3x^2)/2xx1/xdx#

#I=(3x^2)/2lnx-int(3x)/2dx#

#I=(3x^2)/2lnx-(3x^2)/4+C#

#I=(3x^2)/4[2lnx-1]+C#