Integral de $$$\frac{1}{x^{2} - 64}$$$

Encontre $$$\int \frac{1}{x^{2} - 64}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{x^{2} - 64} d x}}} = {\color{red}{\int{\left(- \frac{1}{16 \left(x + 8\right)} + \frac{1}{16 \left(x - 8\right)}\right)d x}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(- \frac{1}{16 \left(x + 8\right)} + \frac{1}{16 \left(x - 8\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{16 \left(x - 8\right)} d x} - \int{\frac{1}{16 \left(x + 8\right)} d x}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{16}$$$ e $$$f{\left(x \right)} = \frac{1}{x + 8}$$$:

$$\int{\frac{1}{16 \left(x - 8\right)} d x} - {\color{red}{\int{\frac{1}{16 \left(x + 8\right)} d x}}} = \int{\frac{1}{16 \left(x - 8\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x + 8} d x}}{16}\right)}}$$

Seja $$$u=x + 8$$$.

Então $$$du=\left(x + 8\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Logo,

$$\int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x + 8} d x}}}}{16} = \int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16}$$

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

$$\int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16} = \int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{16}$$

Recorde que $$$u=x + 8$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{16} + \int{\frac{1}{16 \left(x - 8\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x + 8\right)}}}\right| \right)}}{16} + \int{\frac{1}{16 \left(x - 8\right)} d x}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{16}$$$ e $$$f{\left(x \right)} = \frac{1}{x - 8}$$$:

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + {\color{red}{\int{\frac{1}{16 \left(x - 8\right)} d x}}} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + {\color{red}{\left(\frac{\int{\frac{1}{x - 8} d x}}{16}\right)}}$$

Seja $$$u=x - 8$$$.

Então $$$du=\left(x - 8\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Logo,

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\int{\frac{1}{x - 8} d x}}}}{16} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16}$$

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

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{16}$$

Recorde que $$$u=x - 8$$$:

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{16} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 8\right)}}}\right| \right)}}{16}$$

Portanto,

$$\int{\frac{1}{x^{2} - 64} d x} = \frac{\ln{\left(\left|{x - 8}\right| \right)}}{16} - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16}$$

Adicione a constante de integração:

$$\int{\frac{1}{x^{2} - 64} d x} = \frac{\ln{\left(\left|{x - 8}\right| \right)}}{16} - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16}+C$$

$$$\int \frac{1}{x^{2} - 64}\, dx = \left(\frac{\ln\left(\left|{x - 8}\right|\right)}{16} - \frac{\ln\left(\left|{x + 8}\right|\right)}{16}\right) + C$$$A