Question

How do you evaluate the integral of `int ln(sqrt x)/x`?

Answer 1

Use the fact that `ln(sqrtx) = ln(x^(1/2))=1/2lnx`

Explanation:

So we have:

`int ln(sqrt x)/x = 1/2 int lnx 1/xdx`

Because `d/dx(lnx) = 1/x`, we have

`1/2 int u du = 1/4u^2 +C`

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

Check the answer by differentiating.