Integral de $$$\frac{2 t - 7}{t - 8}$$$

Encontre $$$\int \frac{2 t - 7}{t - 8}\, dt$$$.

Reescreva o numerador do integrando como $$$2 t - 7=2\left(t - 8\right)+9$$$ e decomponha a fração:

$${\color{red}{\int{\frac{2 t - 7}{t - 8} d t}}} = {\color{red}{\int{\left(2 + \frac{9}{t - 8}\right)d t}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(2 + \frac{9}{t - 8}\right)d t}}} = {\color{red}{\left(\int{2 d t} + \int{\frac{9}{t - 8} d t}\right)}}$$

Aplique a regra da constante $$$\int c\, dt = c t$$$ usando $$$c=2$$$:

$$\int{\frac{9}{t - 8} d t} + {\color{red}{\int{2 d t}}} = \int{\frac{9}{t - 8} d t} + {\color{red}{\left(2 t\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=9$$$ e $$$f{\left(t \right)} = \frac{1}{t - 8}$$$:

$$2 t + {\color{red}{\int{\frac{9}{t - 8} d t}}} = 2 t + {\color{red}{\left(9 \int{\frac{1}{t - 8} d t}\right)}}$$

Seja $$$u=t - 8$$$.

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

Assim,

$$2 t + 9 {\color{red}{\int{\frac{1}{t - 8} d t}}} = 2 t + 9 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$2 t + 9 {\color{red}{\int{\frac{1}{u} d u}}} = 2 t + 9 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$2 t + 9 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 2 t + 9 \ln{\left(\left|{{\color{red}{\left(t - 8\right)}}}\right| \right)}$$

Portanto,

$$\int{\frac{2 t - 7}{t - 8} d t} = 2 t + 9 \ln{\left(\left|{t - 8}\right| \right)}$$

Adicione a constante de integração:

$$\int{\frac{2 t - 7}{t - 8} d t} = 2 t + 9 \ln{\left(\left|{t - 8}\right| \right)}+C$$

$$$\int \frac{2 t - 7}{t - 8}\, dt = \left(2 t + 9 \ln\left(\left|{t - 8}\right|\right)\right) + C$$$A