Integral of $$$\frac{\ln\left(y\right)}{y}$$$

Find $$$\int \frac{\ln\left(y\right)}{y}\, dy$$$.

Let $$$u=\ln{\left(y \right)}$$$.

Then $$$du=\left(\ln{\left(y \right)}\right)^{\prime }dy = \frac{dy}{y}$$$ (steps can be seen »), and we have that $$$\frac{dy}{y} = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{\ln{\left(y \right)}}{y} d y}}} = {\color{red}{\int{u d u}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$${\color{red}{\int{u d u}}}={\color{red}{\frac{u^{1 + 1}}{1 + 1}}}={\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

Recall that $$$u=\ln{\left(y \right)}$$$:

$$\frac{{\color{red}{u}}^{2}}{2} = \frac{{\color{red}{\ln{\left(y \right)}}}^{2}}{2}$$

Therefore,

$$\int{\frac{\ln{\left(y \right)}}{y} d y} = \frac{\ln{\left(y \right)}^{2}}{2}$$

Add the constant of integration:

$$\int{\frac{\ln{\left(y \right)}}{y} d y} = \frac{\ln{\left(y \right)}^{2}}{2}+C$$

$$$\int \frac{\ln\left(y\right)}{y}\, dy = \frac{\ln^{2}\left(y\right)}{2} + C$$$A