How do you integrate `int 4 e^x ln x dx` using integration by parts?

Integration by parts is:
`intudv=uv-intvdu`
A helpful acronym for choosing the `u` of IbP is LIATE :
L - logarithmic (ex. `lnx`, `log_3x`)
I - inverse trigonometric (ex. `tan^(-1)x, arcsin(x)`)
A - algebraic (ex. `x^2`, `1/x`)
T - trigonometric (ex. `sinx`, `cosx`)
E - exponential (ex. `e^x`, `2^x`)

In our case, we have a logarithmic (`lnx`) and exponential (`4e^x`) function. In our LIATE list, logarithmic functions come first, so our `u` will be `lnx`. That means everything else, namely `4e^xdx` will be `dv`:
`u=lnx->(du)/dx=1/x->du=1/xdx`
`dv=4e^xdx->intdv=int4e^xdx->v=4e^x`

Applying the parts formula:
`int4e^xlnxdx=(lnx)(4e^x)-int4e^x1/xdx`
`color(white)(XX)=4e^xlnx-4inte^x1/xdx`

We have to do integration by parts a second time for the new integral `4inte^x1/xdx`:
`u=1/x->(du)/dx=-1/x^2->du=-1/x^2dx`
`dv=e^xdx->intdv=inte^xdx->v=e^x`

Applying what we found:
`int4e^xlnxdx=4e^xlnx-4(1/xe^x-inte^x(-1/x^2dx))`
`color(white)(XX)=4e^xlnx-4(e^x/x+inte^x1/x^2dx)`

Ahhhhh! We have to use another integration by parts! But before you do, notice what's going on. Each time we integrate by parts, we end up with something like `e^x1/x` in a new integral. This is, in fact, a pattern. If we keep integrating by parts, we'll get one integral after another, in the form `int(-1)^n(e^x/x^n)dx` - and it will never stop. So this integral continues on forever with no solution.