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

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

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

Then $$$du=\left(\ln{\left(\frac{t}{t + 1} \right)}\right)^{\prime }dt = \frac{1}{t \left(t + 1\right)} dt$$$ (steps can be seen »), and we have that $$$\frac{dt}{t \left(t + 1\right)} = du$$$.

Thus,

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

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

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

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

Therefore,

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

Add the constant of integration:

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

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