Integral dari $$$\frac{6}{\left(3 x - 2\right)^{3}}$$$

Temukan $$$\int \frac{6}{\left(3 x - 2\right)^{3}}\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=6$$$ dan $$$f{\left(x \right)} = \frac{1}{\left(3 x - 2\right)^{3}}$$$:

$${\color{red}{\int{\frac{6}{\left(3 x - 2\right)^{3}} d x}}} = {\color{red}{\left(6 \int{\frac{1}{\left(3 x - 2\right)^{3}} d x}\right)}}$$

Misalkan $$$u=3 x - 2$$$.

Kemudian $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{3}$$$.

Oleh karena itu,

$$6 {\color{red}{\int{\frac{1}{\left(3 x - 2\right)^{3}} d x}}} = 6 {\color{red}{\int{\frac{1}{3 u^{3}} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{3}$$$ dan $$$f{\left(u \right)} = \frac{1}{u^{3}}$$$:

$$6 {\color{red}{\int{\frac{1}{3 u^{3}} d u}}} = 6 {\color{red}{\left(\frac{\int{\frac{1}{u^{3}} d u}}{3}\right)}}$$

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-3$$$:

$$2 {\color{red}{\int{\frac{1}{u^{3}} d u}}}=2 {\color{red}{\int{u^{-3} d u}}}=2 {\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}=2 {\color{red}{\left(- \frac{u^{-2}}{2}\right)}}=2 {\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$

Ingat bahwa $$$u=3 x - 2$$$:

$$- {\color{red}{u}}^{-2} = - {\color{red}{\left(3 x - 2\right)}}^{-2}$$

Oleh karena itu,

$$\int{\frac{6}{\left(3 x - 2\right)^{3}} d x} = - \frac{1}{\left(3 x - 2\right)^{2}}$$

Tambahkan konstanta integrasi:

$$\int{\frac{6}{\left(3 x - 2\right)^{3}} d x} = - \frac{1}{\left(3 x - 2\right)^{2}}+C$$

$$$\int \frac{6}{\left(3 x - 2\right)^{3}}\, dx = - \frac{1}{\left(3 x - 2\right)^{2}} + C$$$A