Evaluate the integral [ln(x)/(x) dx]

Evaluate `int
(lnx)/x dx` :

We ` `` `let `u=lnx` . Then `du=1/xdx` and we have:

`intudu=1/2u^2+C` . Substituting for `u` we get `1/2(lnx)^2+C` .

Thus the integral evaluates as `1/2(ln(x))^2+C`

href="https://en.wikipedia.org/wiki/Logarithm">https://en.wikipedia.org/wiki/Logarithm