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

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

Reescreva e separe a fração:

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=1$$$:

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

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

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

Assim,

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

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

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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