Looking for antiderivativeFind ∫ log ⁡ ( x + 1 + x ) d xI used the subtitution x = tan ⁡ y .but i the integral became too long and involves ∫ sec 3 ⁡ y d yand also ∫ sec 5 ⁡ y d ybut it solvable so i search for a simple solution if exist? Note log is the natural logarithm

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Looking for antiderivativeFind   ∫  log  ⁡      (          x      +      1        +          x        )    d  xI used the subtitution       x    =  tan  ⁡      y  .but i the integral became too long and involves   ∫      sec    3    ⁡      y    d  yand also   ∫      sec    5    ⁡      y    d  ybut it solvable so i search for a simple solution if exist? Note log is the natural logarithm

conveneau71 · Accepted Answer

Explanation:  ∫  ln  ⁡  (      x    +    1    −      x    )  d  x  =  x  ln  ⁡  (      x    +    1    −      x    )  −  ∫            x              (                  1                      2                          x              +              1                                      −                  1                      2                          x                                      )                            x        +        1            −              x              d  x  =  =  x  ln  ⁡  (      x    +    1    −      x    )  +      1    2    ∫            x              x        +        1              d  x  =  .  .  . and substitute             x              x        +        1              =  t  .

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