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

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

Seja $$$u=x + 1$$$.

Então $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Logo,

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

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Recorde que $$$u=x + 1$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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