Integral of $$$\frac{i \left(1 - z\right)}{z + 1}$$$

Find $$$\int \frac{i \left(1 - z\right)}{z + 1}\, dz$$$.

Let $$$u=z + 1$$$.

Then $$$du=\left(z + 1\right)^{\prime }dz = 1 dz$$$ (steps can be seen »), and we have that $$$dz = du$$$.

So,

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

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

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

Expand the expression:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

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

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

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

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

Recall that $$$u=z + 1$$$:

$$i \left(2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - {\color{red}{u}}\right) = i \left(2 \ln{\left(\left|{{\color{red}{\left(z + 1\right)}}}\right| \right)} - {\color{red}{\left(z + 1\right)}}\right)$$

Therefore,

$$\int{\frac{i \left(1 - z\right)}{z + 1} d z} = i \left(- z + 2 \ln{\left(\left|{z + 1}\right| \right)} - 1\right)$$

Add the constant of integration:

$$\int{\frac{i \left(1 - z\right)}{z + 1} d z} = i \left(- z + 2 \ln{\left(\left|{z + 1}\right| \right)} - 1\right)+C$$

$$$\int \frac{i \left(1 - z\right)}{z + 1}\, dz = i \left(- z + 2 \ln\left(\left|{z + 1}\right|\right) - 1\right) + C$$$A