Integral de $$$- x + \frac{x}{a}$$$ em relação a $$$x$$$

Encontre $$$\int \left(- x + \frac{x}{a}\right)\, dx$$$.

Integre termo a termo:

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

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

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

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

$$- \frac{x^{2}}{2} + {\color{red}{\int{\frac{x}{a} d x}}} = - \frac{x^{2}}{2} + {\color{red}{\frac{\int{x d x}}{a}}}$$

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

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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