Question

How do you integrate `[ln sqrt x] / x`?

Answer 1

The integral equals `1/4ln^2x + C`

Explanation:

We can rewrite using logarithm laws.

`I = int ln(x^(1/2))/xdx`

`I =int lnx/(2x) dx`

We now let `u = lnx`. Then `du = 1/xdx` and then `dx= xdu`.

`I = int u/(2x) * x du`

`I = int 1/2u du`

`I = 1/2(1/2u^2) + C`

`I = 1/4ln^2x + C`

Hopefully this helps!