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

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

Untuk integral $$$\int{t \ln{\left(t \right)} d t}$$$, 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 $$$\operatorname{dv}=t dt$$$.

Maka $$$\operatorname{du}=\left(\ln{\left(t \right)}\right)^{\prime }dt=\frac{dt}{t}$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{t d t}=\frac{t^{2}}{2}$$$ (langkah-langkah dapat dilihat di »).

Dengan demikian,

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

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

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

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

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

Oleh karena itu,

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

Sederhanakan:

$$\int{t \ln{\left(t \right)} d t} = \frac{t^{2} \left(2 \ln{\left(t \right)} - 1\right)}{4}$$

Tambahkan konstanta integrasi:

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

$$$\int t \ln\left(t\right)\, dt = \frac{t^{2} \left(2 \ln\left(t\right) - 1\right)}{4} + C$$$A