Integral dari $$$\ln\left(t\right)$$$

Temukan $$$\int \ln\left(t\right)\, dt$$$.

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

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

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

Dengan demikian,

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

Terapkan aturan konstanta $$$\int c\, dt = c t$$$ dengan $$$c=1$$$:

$$t \ln{\left(t \right)} - {\color{red}{\int{1 d t}}} = t \ln{\left(t \right)} - {\color{red}{t}}$$

Oleh karena itu,

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

Sederhanakan:

$$\int{\ln{\left(t \right)} d t} = t \left(\ln{\left(t \right)} - 1\right)$$

Tambahkan konstanta integrasi:

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

$$$\int \ln\left(t\right)\, dt = t \left(\ln\left(t\right) - 1\right) + C$$$A