Can you explain me this antiderivative?Find the antiderivative of e x 2 e x + 2 e x 2 + 5 .The book suggests a switch of variables. Let t = e x 2 . And so x = 2 ln ⁡ ( t ). The antiderivative transforms into: ∫ 2 t 2 + 2 t + 5 d t Here is my first doubt.Where came from the 2 in the numerator?Then, 2 ∫ 1 t 2 + 2 t + 5 d t = 2 ∫ 1 ( t + 1 ) 2 + 4 d t .And finally my main doubt. How can one obtain, 1 2 ∫ 1 1 + ( t + 1 2 ) 2 d tThe antiderivative is tan − 1 ⁡ 1 2 ( e x 2 + 1 ) + C which is different from what Wolfram presents.The Wolfram result was: − tan − 1 ⁡ ( 2 e x 2 + 1 ) + C

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Can you explain me this antiderivative?Find the antiderivative of                     e                              x            2                                                e          x                +        2                  e                                    x              2                                      +        5            .The book suggests a switch of variables. Let   t  =      e                  x        2            . And so   x  =  2  ln  ⁡  (  t  ). The antiderivative transforms into:       ∫          2                        t          2                +        2        t        +        5                    d        t  Here is my first doubt.Where came from the 2 in the numerator?Then,       2    ∫          1                        t          2                +        2        t        +        5                    d        t    =    2    ∫          1              (        t        +        1                  )          2                +        4                    d        t  .And finally my main doubt. How can one obtain,       1    2    ∫      1          1      +      (                        t          +          1                2                    )        2                  d    tThe antiderivative is             tan              −        1              ⁡          1      2        (          e                        x          2                      +    1    )    +    C   which is different from what Wolfram presents.The Wolfram result was:   −      tan          −      1        ⁡  (      2                  e                              x            2                              +      1        )  +  C

Athena Blanchard · Accepted Answer

Step 1First doubt: from   x  =  2  ln  ⁡  (  t  ) you get   d  x  =  2  d  t      /    t, and here comes out the 2.Step 2Second doubt: from   (  t  +  1      )    2    +  4, you get   4  [            (      t      +      1              )        2              4    +  1  ]  =  4  [  (            t      +      1        2        )    2    +  1  ].Step 3Third doubt: take into account the known relation   arctan  ⁡  (  x  )  +  arctan  ⁡  (  1      /    x  )  =  π      /    2 for   x  &amp;gt;  0.

rockandriot0odjn · Accepted Answer

Explanation:  arctan  ⁡      (    1          /        2                                e                          1                      /                    2                    x                      +    1          /        2    )    =  1      /    2    i      (    ln    ⁡          (      −      1              /            2            i                                    e                                1                          /                        2                        x                              +      1      −      1              /            2            i      )        −    ln    ⁡          (      1              /            2            i                                    e                                1                          /                        2                        x                              +      1      +      1              /            2            i      )        )

Can you explain me this antiderivative? Find the antiderivative of <mstyle displaystyle="true"

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