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Integral of lnt/(1+t) dt

Limits of integration:

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Piecewise:

The solution

You have entered [src]
  1          
  /          
 |           
 |  log(t)   
 |  ------ dt
 |  1 + t    
 |           
/            
0            
$$\int\limits_{0}^{1} \frac{\log{\left(t \right)}}{t + 1}\, dt$$
Integral(log(t)/(1 + t), (t, 0, 1))
The answer (Indefinite) [src]
                   //                           -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       |      /                 /
$$\int \frac{\log{\left(t \right)}}{t + 1}\, dt = C + \begin{cases} i \pi \log{\left(t + 1 \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \left|{t + 1}\right| < 1 \\- i \pi \log{\left(\frac{1}{t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \frac{1}{\left|{t + 1}\right|} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left(\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle| {t + 1} \right)} + i \pi {G_{2, 2}^{0, 2}\left(\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle| {t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{otherwise} \end{cases}$$
The answer [src]
    2                
  pi                 
- --- + 2*pi*I*log(2)
   12                
$$- \frac{\pi^{2}}{12} + 2 i \pi \log{\left(2 \right)}$$
=
=
    2                
  pi                 
- --- + 2*pi*I*log(2)
   12                
$$- \frac{\pi^{2}}{12} + 2 i \pi \log{\left(2 \right)}$$
-pi^2/12 + 2*pi*i*log(2)
Numerical answer [src]
-0.822467033424113
-0.822467033424113

    Use the examples entering the upper and lower limits of integration.