Integral of $$$\frac{3}{2 u}$$$

Find $$$\int \frac{3}{2 u}\, du$$$.

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{3}{2}$$$ and $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Therefore,

$$\int{\frac{3}{2 u} d u} = \frac{3 \ln{\left(\left|{u}\right| \right)}}{2}$$

Add the constant of integration:

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

$$$\int \frac{3}{2 u}\, du = \frac{3 \ln\left(\left|{u}\right|\right)}{2} + C$$$A