Integral of $$$\frac{\ln^{2}\left(t\right)}{t}$$$

Find $$$\int \frac{\ln^{2}\left(t\right)}{t}\, dt$$$.

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

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

So,

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$${\color{red}{\int{u^{2} d u}}}={\color{red}{\frac{u^{1 + 2}}{1 + 2}}}={\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$\frac{{\color{red}{u}}^{3}}{3} = \frac{{\color{red}{\ln{\left(t \right)}}}^{3}}{3}$$

Therefore,

$$\int{\frac{\ln{\left(t \right)}^{2}}{t} d t} = \frac{\ln{\left(t \right)}^{3}}{3}$$

Add the constant of integration:

$$\int{\frac{\ln{\left(t \right)}^{2}}{t} d t} = \frac{\ln{\left(t \right)}^{3}}{3}+C$$

$$$\int \frac{\ln^{2}\left(t\right)}{t}\, dt = \frac{\ln^{3}\left(t\right)}{3} + C$$$A