Integraal van $$$\frac{\ln\left(t\right)}{t^{2}}$$$

Bepaal $$$\int \frac{\ln\left(t\right)}{t^{2}}\, dt$$$.

Voor de integraal $$$\int{\frac{\ln{\left(t \right)}}{t^{2}} d t}$$$, gebruik partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Zij $$$\operatorname{u}=\ln{\left(t \right)}$$$ en $$$\operatorname{dv}=\frac{dt}{t^{2}}$$$.

Dan $$$\operatorname{du}=\left(\ln{\left(t \right)}\right)^{\prime }dt=\frac{dt}{t}$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{\frac{1}{t^{2}} d t}=- \frac{1}{t}$$$ (de stappen zijn te zien »).

De integraal wordt

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

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=-1$$$ en $$$f{\left(t \right)} = \frac{1}{t^{2}}$$$:

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

Pas de machtsregel $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

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

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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