How can we evaluate ∫ 0 1 log 2 ⁡ ( x + 1 ) log ⁡ ( x 2 + 1 ) x 2 + 1 d xAny kind of help is appreciated.

Question

How can we evaluate      ∫    0    1                      log        2            ⁡      (      x      +      1      )      log      ⁡              (                  x          2                +        1        )                            x        2            +      1        d  xAny kind of help is appreciated.

allstylekvsvi · Accepted Answer

I found the answer:            ∫      0      1                                log          2                ⁡        (        x        +        1        )        log        ⁡                  (                      x            2                    +          1          )                                      x          2                +        1                  d    x    =                            π          2                C            48        −          15      4        C          log      2        ⁡    (    2    )    +    20    ℑ    (          Li      4        (    1    +    i    )    )    +    2    log    ⁡    (    2    )    ℑ    (          Li      3        (    1    +    i    )    )    +                  35        π        ζ        (        3        )            64        −          5      24        π          log      3        ⁡    (    2    )    −          21      64              π      3        log    ⁡    (    2    )    −          3      256              (              ψ                  (          3          )                            (                  1          4                )            −              ψ                  (          3          )                            (                  3          4                )            )      A generalization:            ∫      0      1                                log          3                ⁡        (        x        +        1        )        log        ⁡                  (                      x            2                    +          1          )                                      x          2                +        1                  d    x    =    −          39      8        C          log      3        ⁡    (    2    )    +          1      32              π      2        C    log    ⁡    (    2    )    −    27    β    (    4    )    log    ⁡    (    2    )    −    48    ℑ          (              Li        5                    (                  1          2                +                  i          2                )            )        −    6          log      2        ⁡    (    2    )    ℑ          (              Li        3                    (                  1          2                +                  i          2                )            )        +    6    log    ⁡    (    2    )    ℑ          (              Li        4                    (                  1          2                +                  i          2                )            )        +          105      128        π    ζ    (    3    )    log    ⁡    (    2    )    +                  119                  π          5                    1024        +          5      16        π          log      4        ⁡    (    2    )    +          105      256              π      3              log      2        ⁡    (    2    )

How can we evaluate <msubsup> ∫ 0 1 </msubsup>

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