Integral de $$$- \frac{x}{2} - 3 \ln\left(- x\right)$$$

Encontre $$$\int \left(- \frac{x}{2} - 3 \ln\left(- x\right)\right)\, dx$$$.

Integre termo a termo:

$${\color{red}{\int{\left(- \frac{x}{2} - 3 \ln{\left(- x \right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{x}{2} d x} - \int{3 \ln{\left(- x \right)} d x}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=3$$$ e $$$f{\left(x \right)} = \ln{\left(- x \right)}$$$:

$$- \int{\frac{x}{2} d x} - {\color{red}{\int{3 \ln{\left(- x \right)} d x}}} = - \int{\frac{x}{2} d x} - {\color{red}{\left(3 \int{\ln{\left(- x \right)} d x}\right)}}$$

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

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

Assim,

$$- \int{\frac{x}{2} d x} - 3 {\color{red}{\int{\ln{\left(- x \right)} d x}}} = - \int{\frac{x}{2} d x} - 3 {\color{red}{\int{\left(- \ln{\left(u \right)}\right)d u}}}$$

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

$$- \int{\frac{x}{2} d x} - 3 {\color{red}{\int{\left(- \ln{\left(u \right)}\right)d u}}} = - \int{\frac{x}{2} d x} - 3 {\color{red}{\left(- \int{\ln{\left(u \right)} d u}\right)}}$$

Para a integral $$$\int{\ln{\left(u \right)} d u}$$$, use integração por partes $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$.

Sejam $$$\operatorname{\kappa}=\ln{\left(u \right)}$$$ e $$$\operatorname{dv}=du$$$.

Então $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{1 d u}=u$$$ (os passos podem ser vistos »).

Logo,

$$- \int{\frac{x}{2} d x} + 3 {\color{red}{\int{\ln{\left(u \right)} d u}}}=- \int{\frac{x}{2} d x} + 3 {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}=- \int{\frac{x}{2} d x} + 3 {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

$$3 u \ln{\left(u \right)} - \int{\frac{x}{2} d x} - 3 {\color{red}{\int{1 d u}}} = 3 u \ln{\left(u \right)} - \int{\frac{x}{2} d x} - 3 {\color{red}{u}}$$

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

$$- \int{\frac{x}{2} d x} - 3 {\color{red}{u}} + 3 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = - \int{\frac{x}{2} d x} - 3 {\color{red}{\left(- x\right)}} + 3 {\color{red}{\left(- x\right)}} \ln{\left({\color{red}{\left(- x\right)}} \right)}$$

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

$$- 3 x \ln{\left(- x \right)} + 3 x - {\color{red}{\int{\frac{x}{2} d x}}} = - 3 x \ln{\left(- x \right)} + 3 x - {\color{red}{\left(\frac{\int{x d x}}{2}\right)}}$$

Aplique a regra da potência $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:

$$- 3 x \ln{\left(- x \right)} + 3 x - \frac{{\color{red}{\int{x d x}}}}{2}=- 3 x \ln{\left(- x \right)} + 3 x - \frac{{\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{2}=- 3 x \ln{\left(- x \right)} + 3 x - \frac{{\color{red}{\left(\frac{x^{2}}{2}\right)}}}{2}$$

Portanto,

$$\int{\left(- \frac{x}{2} - 3 \ln{\left(- x \right)}\right)d x} = - \frac{x^{2}}{4} - 3 x \ln{\left(- x \right)} + 3 x$$

Simplifique:

$$\int{\left(- \frac{x}{2} - 3 \ln{\left(- x \right)}\right)d x} = \frac{x \left(- x - 12 \ln{\left(- x \right)} + 12\right)}{4}$$

Adicione a constante de integração:

$$\int{\left(- \frac{x}{2} - 3 \ln{\left(- x \right)}\right)d x} = \frac{x \left(- x - 12 \ln{\left(- x \right)} + 12\right)}{4}+C$$

$$$\int \left(- \frac{x}{2} - 3 \ln\left(- x\right)\right)\, dx = \frac{x \left(- x - 12 \ln\left(- x\right) + 12\right)}{4} + C$$$A