Integral de $$$\frac{5 - x}{x^{2} - 16}$$$

Encontre $$$\int \frac{5 - x}{x^{2} - 16}\, dx$$$.

Separe a fração:

$${\color{red}{\int{\frac{5 - x}{x^{2} - 16} d x}}} = {\color{red}{\int{\left(- \frac{x}{x^{2} - 16} + \frac{5}{x^{2} - 16}\right)d x}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(- \frac{x}{x^{2} - 16} + \frac{5}{x^{2} - 16}\right)d x}}} = {\color{red}{\left(\int{\left(- \frac{x}{x^{2} - 16}\right)d x} + \int{\frac{5}{x^{2} - 16} d x}\right)}}$$

Seja $$$u=x^{2} - 16$$$.

Então $$$du=\left(x^{2} - 16\right)^{\prime }dx = 2 x dx$$$ (veja os passos »), e obtemos $$$x dx = \frac{du}{2}$$$.

Logo,

$$\int{\frac{5}{x^{2} - 16} d x} + {\color{red}{\int{\left(- \frac{x}{x^{2} - 16}\right)d x}}} = \int{\frac{5}{x^{2} - 16} d x} + {\color{red}{\int{\left(- \frac{1}{2 u}\right)d u}}}$$

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

$$\int{\frac{5}{x^{2} - 16} d x} + {\color{red}{\int{\left(- \frac{1}{2 u}\right)d u}}} = \int{\frac{5}{x^{2} - 16} d x} + {\color{red}{\left(- \frac{\int{\frac{1}{u} d u}}{2}\right)}}$$

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

$$\int{\frac{5}{x^{2} - 16} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \int{\frac{5}{x^{2} - 16} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Recorde que $$$u=x^{2} - 16$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + \int{\frac{5}{x^{2} - 16} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x^{2} - 16\right)}}}\right| \right)}}{2} + \int{\frac{5}{x^{2} - 16} d x}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=5$$$ e $$$f{\left(x \right)} = \frac{1}{x^{2} - 16}$$$:

$$- \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + {\color{red}{\int{\frac{5}{x^{2} - 16} d x}}} = - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + {\color{red}{\left(5 \int{\frac{1}{x^{2} - 16} d x}\right)}}$$

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

$$- \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 {\color{red}{\int{\frac{1}{x^{2} - 16} d x}}} = - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 {\color{red}{\int{\left(- \frac{1}{8 \left(x + 4\right)} + \frac{1}{8 \left(x - 4\right)}\right)d x}}}$$

Integre termo a termo:

$$- \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 {\color{red}{\int{\left(- \frac{1}{8 \left(x + 4\right)} + \frac{1}{8 \left(x - 4\right)}\right)d x}}} = - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 {\color{red}{\left(\int{\frac{1}{8 \left(x - 4\right)} d x} - \int{\frac{1}{8 \left(x + 4\right)} d x}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{8}$$$ e $$$f{\left(x \right)} = \frac{1}{x + 4}$$$:

$$- \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x} - 5 {\color{red}{\int{\frac{1}{8 \left(x + 4\right)} d x}}} = - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x} - 5 {\color{red}{\left(\frac{\int{\frac{1}{x + 4} d x}}{8}\right)}}$$

Seja $$$u=x + 4$$$.

Então $$$du=\left(x + 4\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

A integral torna-se

$$- \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{5 {\color{red}{\int{\frac{1}{x + 4} d x}}}}{8} = - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

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

$$- \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{8} = - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Recorde que $$$u=x + 4$$$:

$$- \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} - \frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x} = - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} - \frac{5 \ln{\left(\left|{{\color{red}{\left(x + 4\right)}}}\right| \right)}}{8} + 5 \int{\frac{1}{8 \left(x - 4\right)} d x}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{8}$$$ e $$$f{\left(x \right)} = \frac{1}{x - 4}$$$:

$$- \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 {\color{red}{\int{\frac{1}{8 \left(x - 4\right)} d x}}} = - \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + 5 {\color{red}{\left(\frac{\int{\frac{1}{x - 4} d x}}{8}\right)}}$$

Seja $$$u=x - 4$$$.

Então $$$du=\left(x - 4\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Portanto,

$$- \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + \frac{5 {\color{red}{\int{\frac{1}{x - 4} d x}}}}{8} = - \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

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

$$- \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{8} = - \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Recorde que $$$u=x - 4$$$:

$$- \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + \frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} = - \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2} + \frac{5 \ln{\left(\left|{{\color{red}{\left(x - 4\right)}}}\right| \right)}}{8}$$

Portanto,

$$\int{\frac{5 - x}{x^{2} - 16} d x} = \frac{5 \ln{\left(\left|{x - 4}\right| \right)}}{8} - \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2}$$

Adicione a constante de integração:

$$\int{\frac{5 - x}{x^{2} - 16} d x} = \frac{5 \ln{\left(\left|{x - 4}\right| \right)}}{8} - \frac{5 \ln{\left(\left|{x + 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x^{2} - 16}\right| \right)}}{2}+C$$

$$$\int \frac{5 - x}{x^{2} - 16}\, dx = \left(\frac{5 \ln\left(\left|{x - 4}\right|\right)}{8} - \frac{5 \ln\left(\left|{x + 4}\right|\right)}{8} - \frac{\ln\left(\left|{x^{2} - 16}\right|\right)}{2}\right) + C$$$A