To find the integral of \(\frac{\ln y}{y}\), we can use the substitution method. Let's set:
\[ u = \ln y \]
Then, we find the differential of \(u\):
\[ du = \frac{1}{y} dy \quad \text{or} \quad dy = y , du = e^u , du \]
Substituting \(y = e^u\) into the integral, we have:
\[ \int \frac{\ln y}{y} , dy = \int u , du \]
Now, we can integrate \(u\):
\[ \int u , du = \frac{u^2}{2} + C \]
Substituting back \(u = \ln y\):
\[ \int \frac{\ln y}{y} , dy = \frac{(\ln y)^2}{2} + C \]
Thus, the integral of \(\frac{\ln y}{y}\) is:
\[ \int \frac{\ln y}{y} , dy = \frac{(\ln y)^2}{2} + C \]
where \(C\) is the constant of integration.