Integral dari $$$- \frac{1}{2 x - 5} + \frac{1}{2 x^{5}}$$$

Temukan $$$\int \left(- \frac{1}{2 x - 5} + \frac{1}{2 x^{5}}\right)\, dx$$$.

Integralkan suku demi suku:

$${\color{red}{\int{\left(- \frac{1}{2 x - 5} + \frac{1}{2 x^{5}}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{2 x^{5}} d x} - \int{\frac{1}{2 x - 5} d x}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(x \right)} = \frac{1}{x^{5}}$$$:

$$- \int{\frac{1}{2 x - 5} d x} + {\color{red}{\int{\frac{1}{2 x^{5}} d x}}} = - \int{\frac{1}{2 x - 5} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x^{5}} d x}}{2}\right)}}$$

Terapkan aturan pangkat $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-5$$$:

$$- \int{\frac{1}{2 x - 5} d x} + \frac{{\color{red}{\int{\frac{1}{x^{5}} d x}}}}{2}=- \int{\frac{1}{2 x - 5} d x} + \frac{{\color{red}{\int{x^{-5} d x}}}}{2}=- \int{\frac{1}{2 x - 5} d x} + \frac{{\color{red}{\frac{x^{-5 + 1}}{-5 + 1}}}}{2}=- \int{\frac{1}{2 x - 5} d x} + \frac{{\color{red}{\left(- \frac{x^{-4}}{4}\right)}}}{2}=- \int{\frac{1}{2 x - 5} d x} + \frac{{\color{red}{\left(- \frac{1}{4 x^{4}}\right)}}}{2}$$

Misalkan $$$u=2 x - 5$$$.

Kemudian $$$du=\left(2 x - 5\right)^{\prime }dx = 2 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{2}$$$.

Dengan demikian,

$$- {\color{red}{\int{\frac{1}{2 x - 5} d x}}} - \frac{1}{8 x^{4}} = - {\color{red}{\int{\frac{1}{2 u} d u}}} - \frac{1}{8 x^{4}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$- {\color{red}{\int{\frac{1}{2 u} d u}}} - \frac{1}{8 x^{4}} = - {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}} - \frac{1}{8 x^{4}}$$

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

$$- \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} - \frac{1}{8 x^{4}} = - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2} - \frac{1}{8 x^{4}}$$

Ingat bahwa $$$u=2 x - 5$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \frac{1}{8 x^{4}} = - \frac{\ln{\left(\left|{{\color{red}{\left(2 x - 5\right)}}}\right| \right)}}{2} - \frac{1}{8 x^{4}}$$

Oleh karena itu,

$$\int{\left(- \frac{1}{2 x - 5} + \frac{1}{2 x^{5}}\right)d x} = - \frac{\ln{\left(\left|{2 x - 5}\right| \right)}}{2} - \frac{1}{8 x^{4}}$$

Tambahkan konstanta integrasi:

$$\int{\left(- \frac{1}{2 x - 5} + \frac{1}{2 x^{5}}\right)d x} = - \frac{\ln{\left(\left|{2 x - 5}\right| \right)}}{2} - \frac{1}{8 x^{4}}+C$$

$$$\int \left(- \frac{1}{2 x - 5} + \frac{1}{2 x^{5}}\right)\, dx = \left(- \frac{\ln\left(\left|{2 x - 5}\right|\right)}{2} - \frac{1}{8 x^{4}}\right) + C$$$A