Integral dari $$$\ln\left(2 n\right)$$$

Temukan $$$\int \ln\left(2 n\right)\, dn$$$.

Misalkan $$$u=2 n$$$.

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

Integralnya menjadi

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

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

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

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

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

Maka $$$\operatorname{dg}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (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 $$$u=2 n$$$:

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

Oleh karena itu,

$$\int{\ln{\left(2 n \right)} d n} = n \ln{\left(2 n \right)} - n$$

Sederhanakan:

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

Tambahkan konstanta integrasi:

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

$$$\int \ln\left(2 n\right)\, dn = n \left(\ln\left(n\right) - 1 + \ln\left(2\right)\right) + C$$$A