Integral de $$$\frac{1}{4 - x^{2}}$$$

Encontre $$$\int \frac{1}{4 - x^{2}}\, dx$$$.

Efetue a decomposição em frações parciais (os passos podem ser vistos »):

$${\color{red}{\int{\frac{1}{4 - x^{2}} d x}}} = {\color{red}{\int{\left(\frac{1}{4 \left(x + 2\right)} - \frac{1}{4 \left(x - 2\right)}\right)d x}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(\frac{1}{4 \left(x + 2\right)} - \frac{1}{4 \left(x - 2\right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{4 \left(x - 2\right)} d x} + \int{\frac{1}{4 \left(x + 2\right)} d x}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{4}$$$ e $$$f{\left(x \right)} = \frac{1}{x - 2}$$$:

$$\int{\frac{1}{4 \left(x + 2\right)} d x} - {\color{red}{\int{\frac{1}{4 \left(x - 2\right)} d x}}} = \int{\frac{1}{4 \left(x + 2\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x - 2} d x}}{4}\right)}}$$

Seja $$$u=x - 2$$$.

Então $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

A integral pode ser reescrita como

$$\int{\frac{1}{4 \left(x + 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x - 2} d x}}}}{4} = \int{\frac{1}{4 \left(x + 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4}$$

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

$$\int{\frac{1}{4 \left(x + 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4} = \int{\frac{1}{4 \left(x + 2\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

Recorde que $$$u=x - 2$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} + \int{\frac{1}{4 \left(x + 2\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}}{4} + \int{\frac{1}{4 \left(x + 2\right)} d x}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{4}$$$ e $$$f{\left(x \right)} = \frac{1}{x + 2}$$$:

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + {\color{red}{\int{\frac{1}{4 \left(x + 2\right)} d x}}} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + {\color{red}{\left(\frac{\int{\frac{1}{x + 2} d x}}{4}\right)}}$$

Seja $$$u=x + 2$$$.

Então $$$du=\left(x + 2\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Assim,

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + \frac{{\color{red}{\int{\frac{1}{x + 2} d x}}}}{4} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4}$$

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

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

Recorde que $$$u=x + 2$$$:

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + \frac{\ln{\left(\left|{{\color{red}{\left(x + 2\right)}}}\right| \right)}}{4}$$

Portanto,

$$\int{\frac{1}{4 - x^{2}} d x} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{4} + \frac{\ln{\left(\left|{x + 2}\right| \right)}}{4}$$

Simplifique:

$$\int{\frac{1}{4 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 2}\right| \right)} + \ln{\left(\left|{x + 2}\right| \right)}}{4}$$

Adicione a constante de integração:

$$\int{\frac{1}{4 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 2}\right| \right)} + \ln{\left(\left|{x + 2}\right| \right)}}{4}+C$$

$$$\int \frac{1}{4 - x^{2}}\, dx = \frac{- \ln\left(\left|{x - 2}\right|\right) + \ln\left(\left|{x + 2}\right|\right)}{4} + C$$$A