Answer
$$\int \frac{\ln\sqrt t}{t}dt=\frac{\ln^2t}{4}+C$$
Work Step by Step
$$A=\int \frac{\ln\sqrt t}{t}dt=\int\frac{\ln t^{1/2}}{t}dt$$ $$A=\frac{1}{2}\int\frac{\ln t}{t}dt$$
We set $u=\ln t$
Then $$du=\frac{1}{t}dt$$
Therefore, $$A=\frac{1}{2}\int udu=\frac{1}{2}\times\frac{u^2}{2}+C$$ $$A=\frac{u^2}{4}+C$$ $$A=\frac{\ln^2t}{4}+C$$