A closed form of ∫ 0 1 arctan ⁡ ( x 3 ) 1 + x d x ?

Question

A closed form of       ∫    0    1              arctan      ⁡      (              x        3            )              1      +      x        d  x ?

Jake Mcpherson · Accepted Answer

Firstly, set this integral as  I  =      ∫    0    1              arctan      ⁡              (                  x          3                )                    1      +      x                 d    x  .Then, we have                    I                            =                  ∫          0          1                                      arctan            ⁡                          (                              x                3                            )                                            1            +            x                                               d                x                                          =                  ∫          0          1                arctan        ⁡                  (                      x            3                    )                                     d                ln        ⁡        (        1        +        x        )                                          =                              π            ln            ⁡            2                    4                −        3                  ∫          0          1                                                    x              2                        ln            ⁡            (            1            +            x            )                                1            +                          x              6                                                           d                x                                          =                              π            ln            ⁡            2                    4                +                              ∫            0            1                                              ln              ⁡              (              1              +              x              )                                      1              +                              x                2                                                                       d                    x                                                          −                  ∫          0          1                                                    (              1              +                              x                2                            )                        ln            ⁡            (            1            +            x            )                                1            −                          x              2                        +                          x              4                                                           d                x                                          =                              π            ln            ⁡            2                    4                +                                            π              ln              ⁡              2                        8                          −                  1          2                          ∫          0          1                                      ln            ⁡            (            1            +            x            )                                1            −                          3                        x            +                          x              2                                                           d                x                                                  −                  1          2                          ∫          0          1                                      ln            ⁡            (            1            +            x            )                                1            +                          3                        x            +                          x              2                                                           d                x                                          =                              3            π            ln            ⁡            2                    8                −                  1          2                (                  J          1                +                  J          2                )        .            Where the red integral could be calculated as                                                                                                        ∫                  0                  1                                                                      ln                    ⁡                    (                    1                    +                    x                    )                                                        1                    +                                          x                      2                                                                                                           d                                x                            ⏟                                            x            →                                          1                −                x                                            1                +                x                                                                                      =                  ∫          0          1                                      ln            ⁡            2            −            ln            ⁡            (            1            +            x            )                                1            +                          x              2                                                           d                x                                          =                              ln            ⁡            2                    2                          ∫          0          1                          1                      1            +                          x              2                                                           d                x                                          =                              π            ln            ⁡            2                    8                .            Next, we might as well let      J    1    (  a  )  =      ∫    0    1              ln      ⁡      (      1      +      a      x      )              1      −              3            x      +              x        2                       d    x  .Where   a  &amp;gt;  −  1. Then, we get                              J          1          ′                (        a        )                            =                  ∫          0          1                          x                      (            1            +            a            x            )                          (              1              −                              3                            x              +                              x                2                            )                                                           d                x                                          =                  1                      2                          (              1              +                              3                            a              +                              a                2                            )                                                ∫          0          1                                      2            x            −                          3                                            1            −                          3                        x            +                          x              2                                                           d                x                                                  −                  a                      1            +                          3                        a            +                          a              2                                                ∫          0          1                          1                      1            +            a            x                                               d                x                                                  +                              2            a            +                          3                                            2                          (              1              +                              3                            a              +                              a                2                            )                                                ∫          0          1                          1                      1            −                          3                        x            +                          x              2                                                           d                x                                          =                                                        1                              2                                  (                  1                  +                                      3                                    a                  +                                      a                    2                                    )                                                      ln            ⁡                          [                                                1                  −                                      3                                    x                  +                                      x                    2                                                                    (                  1                  +                  a                  x                                      )                    2                                                              ]                        |                    0          1                                                          +                              2            (            2            a            +                          3                                    )                                1            +                          3                        a            +                          a              2                                                ∫          0          1                          1                      1            +            (            2            x            −                          3                                                  )              2                                                           d                x                                          =                              ln            ⁡            (            2            −                          3                                    )            −            2            ln            ⁡            (            1            +            a            )                                2                          (              1              +                              3                            a              +                              a                2                            )                                                                                +                              2            a            +                          3                                            1            +                          3                        a            +                          a              2                                                                        arctan            ⁡            (            2            x            −                          3                                    )            |                    0          1                                                  =                              ln            ⁡            (            2            −                          3                                    )                                2                          (              1              +                              3                            a              +                              a                2                            )                                      +                              5            π            (            2            a            +                          3                                    )                                12                          (              1              +                              3                            a              +                              a                2                            )                                                                                −                              ln            ⁡            (            1            +            a            )                                1            +                          3                        a            +                          a              2                                      .            Note that       J    1    (  0  )  =  0,       J    1    =      J    1    (  1  ), hence                              J          1                                    =                  J          1                (        1        )        =                  ∫          0          1                          J          1          ′                (        a        )                             d                a        +                  J          1                (        0        )                                          =                              ln            ⁡            (            2            −                          3                                    )                    2                          ∫          0          1                          1                      1            +                          3                        a            +                          a              2                                                           d                a                                                  +                              5            π                    12                          ∫          0          1                                      2            a            +                          3                                            1            +                          3                        a            +                          a              2                                                           d                a                                                  −                                                                              ∫                  0                  1                                                                      ln                    ⁡                    (                    1                    +                    a                    )                                                        1                    +                                          3                                        a                    +                                          a                      2                                                                                                           d                                a                            ⏟                                            a            →            x                                                            =        2        ln        ⁡        (        2        −                  3                        )                  ∫          0          1                          1                      1            +            (            2            a            +                          3                                                  )              2                                                           d                a                                                  +                              5            π                    12                ln        ⁡                  (          1          +                      3                    a          +                      a            2                    )                                                  |                    0          1                                                          −                  ∫          0          1                                      ln            ⁡            (            1            +            x            )                                1            +                          3                        x            +                          x              2                                                           d                x                                          =                              π            ln            ⁡            (            2            −                          3                                    )            +            5            π            ln            ⁡            (            2            +                          3                                    )                    12                                                          −                  ∫          0          1                                      ln            ⁡            (            1            +            x            )                                1            +                          3                        x            +                          x              2                                                           d                x                                          =                              π            ln            ⁡            (            2            +                          3                                    )                    3                −                  ∫          0          1                                      ln            ⁡            (            1            +            x            )                                1            +                          3                        x            +                          x              2                                                           d                x        .            After that, note that      J    1    (  1  )  =      ∫    0    1              ln      ⁡      (      1      +      x      )              1      −              3            x      +              x        2                       d    x      J    2    =      ∫    0    1              ln      ⁡      (      1      +      x      )              1      +              3            x      +              x        2                       d    x  .Thus                                                    ∫          0          1                                      ln            ⁡            (            1            +            x            )                                1            −                          3                        x            +                          x              2                                                           d                x        +                  ∫          0          1                                      ln            ⁡            (            1            +            x            )                                1            +                          3                        x            +                          x              2                                                           d                x                                          =                  J          1                +                  J          2                =                  ∫          0          1                                                    (              1              +                              x                2                            )                        ln            ⁡            (            1            +            x            )                                1            −                          x              2                        +                          x              4                                                           d                x                                          =                              π            ln            ⁡            (            2            +                          3                                    )                    3                .            Finally, we obtain                    I                            =                              3            π            ln            ⁡            2                    8                −                  1          2                (                  J          1                +                  J          2                )                                          =                              3            π            ln            ⁡            2                    8                −                  π          6                ln        ⁡        (        2        +                  3                        )        .

A closed form of <msubsup> ∫ 0 1 </msubsup> arctan

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