How do you evaluate the integral $int xlnxdx$ ?

Question

2584 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-10T11:11:14

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

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.

$int xlnxdx$

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

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

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

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

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

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

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

How do you evaluate the integral int xlnxdxint xlnxdx?