Integral of lnt/(1+t) dt

                   //                           -polylog(2, 1 + t) + pi*I*log(1 + t)                             for |1 + t| < 1\
  /                ||                                                                                                           |
 |                 ||                                                        /  1  \                                    1       |
 | log(t)          ||                           -polylog(2, 1 + t) - pi*I*log|-----|                             for ------- < 1|
 | ------ dt = C + |<                                                        \1 + t/                                 |1 + t|    |
 | 1 + t           ||                                                                                                           |
 |                 ||                           __0, 2 /1, 1       |      \         __2, 0 /      1, 1 |      \                 |
/                  ||-polylog(2, 1 + t) + pi*I*/__     |           | 1 + t| - pi*I*/__     |           | 1 + t|     otherwise   |
                   \\                          \_|2, 2 \      0, 0 |      /        \_|2, 2 \0, 0       |      /                 /

$$\int \frac{\log{\left(t \right)}}{t + 1}\, dt = C + \begin{cases} i \pi \log{\left(t + 1 \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \left|{t + 1}\right| < 1 \\- i \pi \log{\left(\frac{1}{t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \frac{1}{\left|{t + 1}\right|} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left(\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle| {t + 1} \right)} + i \pi {G_{2, 2}^{0, 2}\left(\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle| {t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{otherwise} \end{cases}$$

Simplify