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

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

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

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

The integral becomes

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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