Integral dari $$$- \frac{2}{\left(2 x - 9\right)^{2}}$$$

Temukan $$$\int \left(- \frac{2}{\left(2 x - 9\right)^{2}}\right)\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=-2$$$ dan $$$f{\left(x \right)} = \frac{1}{\left(2 x - 9\right)^{2}}$$$:

$${\color{red}{\int{\left(- \frac{2}{\left(2 x - 9\right)^{2}}\right)d x}}} = {\color{red}{\left(- 2 \int{\frac{1}{\left(2 x - 9\right)^{2}} d x}\right)}}$$

Misalkan $$$u=2 x - 9$$$.

Kemudian $$$du=\left(2 x - 9\right)^{\prime }dx = 2 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{2}$$$.

Dengan demikian,

$$- 2 {\color{red}{\int{\frac{1}{\left(2 x - 9\right)^{2}} d x}}} = - 2 {\color{red}{\int{\frac{1}{2 u^{2}} d u}}}$$

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

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

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

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

Ingat bahwa $$$u=2 x - 9$$$:

$${\color{red}{u}}^{-1} = {\color{red}{\left(2 x - 9\right)}}^{-1}$$

Oleh karena itu,

$$\int{\left(- \frac{2}{\left(2 x - 9\right)^{2}}\right)d x} = \frac{1}{2 x - 9}$$

Tambahkan konstanta integrasi:

$$\int{\left(- \frac{2}{\left(2 x - 9\right)^{2}}\right)d x} = \frac{1}{2 x - 9}+C$$

$$$\int \left(- \frac{2}{\left(2 x - 9\right)^{2}}\right)\, dx = \frac{1}{2 x - 9} + C$$$A