Integral dari $$$x + 1 - \frac{1}{x}$$$

Temukan $$$\int \left(x + 1 - \frac{1}{x}\right)\, dx$$$.

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=1$$$:

$$- \int{\frac{1}{x} d x} + \int{x d x} + {\color{red}{\int{1 d x}}} = - \int{\frac{1}{x} d x} + \int{x d x} + {\color{red}{x}}$$

Terapkan aturan pangkat $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=1$$$:

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

Integral dari $$$\frac{1}{x}$$$ adalah $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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