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

Find $$$\int \ln\left(d\right)\, dd$$$.

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

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

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

So,

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

Apply the constant rule $$$\int c\, dd = c d$$$ with $$$c=1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \ln\left(d\right)\, dd = d \left(\ln\left(d\right) - 1\right) + C$$$A