Question

How do you evaluate the integral `int xlnxdx`?

Answer 1

`1/4x^2(2lnx-1)+C`

Explanation:

This needs the method of 'integration by parts'. The formula of which needs to be known.

`intu(dv)/(dx)dx=uv-intv(du)/(dx)dx`

Care must be taken in the choice of `u` & `(dv)/(dx)` otherwise you end up with a more difficult, or impossible integration!

When logs are involved take the `u` as the log part, the reason will become clear as we work through this problem.

`int xlnxdx`

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

`(dv)/(dx)=x=>v=1/2x^2`

`I=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx`

`I=1/2x^2lnx-int1/2x^2xx 1/xdx`

`I=1/2x^2lnx-1/2intxdx`

`I=1/2x^2lnx-1/4x^2 +C`

`I=1/4x^2(2lnx-1)+C`