Question

How do I find the integral `intln(x)/sqrt(x)dx` ?

Answer 1

You can proceed by parts , that is, using the formula
`\int u dv=uv-\int v du`. See integration by parts

This means that you need to pick what you will interpret as `u` (to be differentiated) and what you will interpret as `dv` (to be integrated).

Since you do not know (or, I should say, are usually not expected to know but rather to know how to obtain: by parts) an antiderivative of `ln x` but you know its derivative, you have no choice: you differentiate `ln x` and integrate `1/sqrt x=x^(-1/2)`.

In other words,
`u=ln x` so that `du=dx/x` and `dv=x^-(1/2)` so that `v=\int x^(-1/2)dx=2 x^(1/2)`. Applying the formula above
`\int \ln x/sqrt x dx=2sqrt x ln x-2\int x^(1/2)/x dx=2sqrt x ln x-2\int x^(-1/2) dx` so that
`\int \ln x/sqrt x dx=2sqrt x ln x-4sqrt x+C`