Question

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

Answer 1

`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`