Integral dari $$$\ln\left(4 - 2 x\right)$$$

Temukan $$$\int \ln\left(4 - 2 x\right)\, dx$$$.

Misalkan $$$u=4 - 2 x$$$.

Kemudian $$$du=\left(4 - 2 x\right)^{\prime }dx = - 2 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = - \frac{du}{2}$$$.

Jadi,

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

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

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

Untuk integral $$$\int{\ln{\left(u \right)} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$.

Misalkan $$$\operatorname{\kappa}=\ln{\left(u \right)}$$$ dan $$$\operatorname{dv}=du$$$.

Maka $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{1 d u}=u$$$ (langkah-langkah dapat dilihat di »).

Jadi,

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

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:

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

Ingat bahwa $$$u=4 - 2 x$$$:

$$\frac{{\color{red}{u}}}{2} - \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{2} = \frac{{\color{red}{\left(4 - 2 x\right)}}}{2} - \frac{{\color{red}{\left(4 - 2 x\right)}} \ln{\left({\color{red}{\left(4 - 2 x\right)}} \right)}}{2}$$

Oleh karena itu,

$$\int{\ln{\left(4 - 2 x \right)} d x} = - x - \frac{\left(4 - 2 x\right) \ln{\left(4 - 2 x \right)}}{2} + 2$$

Sederhanakan:

$$\int{\ln{\left(4 - 2 x \right)} d x} = \left(x - 2\right) \left(\ln{\left(2 - x \right)} - 1 + \ln{\left(2 \right)}\right)$$

Tambahkan konstanta integrasi:

$$\int{\ln{\left(4 - 2 x \right)} d x} = \left(x - 2\right) \left(\ln{\left(2 - x \right)} - 1 + \ln{\left(2 \right)}\right)+C$$

$$$\int \ln\left(4 - 2 x\right)\, dx = \left(x - 2\right) \left(\ln\left(2 - x\right) - 1 + \ln\left(2\right)\right) + C$$$A