Integral dari $$$a^{2} + \frac{1}{x^{2}}$$$ terhadap $$$x$$$

Temukan $$$\int \left(a^{2} + \frac{1}{x^{2}}\right)\, dx$$$.

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=a^{2}$$$:

$$\int{\frac{1}{x^{2}} d x} + {\color{red}{\int{a^{2} d x}}} = \int{\frac{1}{x^{2}} d x} + {\color{red}{a^{2} x}}$$

Terapkan aturan pangkat $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-2$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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