Integral of $$$\frac{z}{z - \frac{3}{2}}$$$

Find $$$\int \frac{z}{z - \frac{3}{2}}\, dz$$$.

Simplify:

$${\color{red}{\int{\frac{z}{z - \frac{3}{2}} d z}}} = {\color{red}{\int{\frac{2 z}{2 z - 3} d z}}}$$

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

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

Rewrite the numerator of the integrand as $$$z=\frac{1}{2}\left(2 z - 3\right)+\frac{3}{2}$$$ and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dz = c z$$$ with $$$c=\frac{1}{2}$$$:

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

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

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

Let $$$u=2 z - 3$$$.

Then $$$du=\left(2 z - 3\right)^{\prime }dz = 2 dz$$$ (steps can be seen »), and we have that $$$dz = \frac{du}{2}$$$.

Thus,

$$z + 3 {\color{red}{\int{\frac{1}{2 z - 3} d z}}} = z + 3 {\color{red}{\int{\frac{1}{2 u} d u}}}$$

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

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

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

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

Recall that $$$u=2 z - 3$$$:

$$z + \frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = z + \frac{3 \ln{\left(\left|{{\color{red}{\left(2 z - 3\right)}}}\right| \right)}}{2}$$

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{z}{z - \frac{3}{2}}\, dz = \left(z + \frac{3 \ln\left(\left|{2 z - 3}\right|\right)}{2}\right) + C$$$A