Question

What is the antiderivative of `x ln x`?

Answer 1

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

Explanation:

Using integration by parts:

`intudv=uv-intvdu`

In the case of

`intxlnxdx`

We let

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

`dv=xdx" "=>" "intdv=intxdx" "=>" "v=x^2/2`

Thus, plugging these in, we see that

`intxlnxdx=(x^2/2)lnx-intx^2/2(1/x)dx`

`=(x^2lnx)/2-intx/2dx`

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