Integral of $$$\frac{x - 1}{x + 1}$$$

Find $$$\int \frac{x - 1}{x + 1}\, dx$$$.

Let $$$u=x + 1$$$.

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

The integral becomes

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

Expand the expression:

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

Integrate term by term:

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

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

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration (and remove the constant from the expression):

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

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