Answer
$$\int\frac{dy}{y\sqrt{3+(\ln y)^2}}=\sinh^{-1}\frac{\ln y}{\sqrt3}+C$$
Work Step by Step
$$A=\int\frac{dy}{y\sqrt{3+(\ln y)^2}}$$
We set $u=\ln y$, which means $$du=\frac{dy}{y}$$
Therefore, $$A=\int\frac{du}{\sqrt{3+u^2}}$$
Use Formula 34, which states that
$$\int\frac{dx}{\sqrt{a^2+x^2}}=\sinh^{-1}\frac{x}{a}+C$$
for $a=\sqrt3$.
$$A=\sinh^{-1}\frac{u}{\sqrt3}+C$$ $$A=\sinh^{-1}\frac{\ln y}{\sqrt3}+C$$
