Integral of $$$\ln\left(t\right)$$$

Find $$$\int \ln\left(t\right)\, dt$$$.

For the integral $$$\int{\ln{\left(t \right)} d t}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\ln{\left(t \right)}$$$ and $$$\operatorname{dv}=dt$$$.

Then $$$\operatorname{du}=\left(\ln{\left(t \right)}\right)^{\prime }dt=\frac{dt}{t}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d t}=t$$$ (steps can be seen »).

The integral becomes

$${\color{red}{\int{\ln{\left(t \right)} d t}}}={\color{red}{\left(\ln{\left(t \right)} \cdot t-\int{t \cdot \frac{1}{t} d t}\right)}}={\color{red}{\left(t \ln{\left(t \right)} - \int{1 d t}\right)}}$$

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=1$$$:

$$t \ln{\left(t \right)} - {\color{red}{\int{1 d t}}} = t \ln{\left(t \right)} - {\color{red}{t}}$$

Therefore,

$$\int{\ln{\left(t \right)} d t} = t \ln{\left(t \right)} - t$$

Simplify:

$$\int{\ln{\left(t \right)} d t} = t \left(\ln{\left(t \right)} - 1\right)$$

Add the constant of integration:

$$\int{\ln{\left(t \right)} d t} = t \left(\ln{\left(t \right)} - 1\right)+C$$

$$$\int \ln\left(t\right)\, dt = t \left(\ln\left(t\right) - 1\right) + C$$$A