Integral de $$$\frac{\ln\left(\frac{t}{t + 1}\right)}{t \left(t + 1\right)}$$$

Encontre $$$\int \frac{\ln\left(\frac{t}{t + 1}\right)}{t \left(t + 1\right)}\, dt$$$.

Seja $$$u=\ln{\left(\frac{t}{t + 1} \right)}$$$.

Então $$$du=\left(\ln{\left(\frac{t}{t + 1} \right)}\right)^{\prime }dt = \frac{1}{t \left(t + 1\right)} dt$$$ (veja os passos »), e obtemos $$$\frac{dt}{t \left(t + 1\right)} = du$$$.

A integral pode ser reescrita como

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

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:

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

Recorde que $$$u=\ln{\left(\frac{t}{t + 1} \right)}$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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