Antiderivatives such as ∫ d x e 2 x + c

Question

Antiderivatives such as   ∫            d      x                      e                  2          x                    +      c

Jamarcus Shields · Accepted Answer

Step 1Firstly, the integral   ∫      1                  u        2            −              c        2              d  u  =  −      1    c    artanh  ⁡      u    c    +  C is quite standard, so really if you&#039;d have known it your method would be pretty efficient!Step 2Secondly, you could first make the substitution   u  =      e    x        /              c      , so that       e    x    =  u      c   and   d  u      /    d  x  =      e    x        /        c    =  u    ⟹    d  x  =  d  u      /    u. This means that if I is the integral then                     I                            =        ∫                  1                                    e                              2                x                                      +            c                          d        x                                          =        ∫                  1                                    u              2                        c            +            c                          ×                  1          u                d        u                                          =                  1                      c                          ∫                  1                      u                                          u                2                            +              1                                      d        u            Here we have a not-very-standard integral:   ∫      1          u                        u          2                +        1              d  u  =  −  arcosech  ⁡  u  +  C  =  −  arsinh  ⁡      1    u    +  C

ntaraxq · Accepted Answer

Step 1You can write this integral as   I  =  ∫                    e                  −          x                    d      x              1      +      c              e                  −          2          x                    and make the substitution (for   c  &amp;gt;  0):      c        e          −      x        =  sinh  ⁡  t      1    +    c          e              −        2        x              =  cosh  ⁡  t  −      c        e          −      x        d  x  =  cosh  ⁡  t    d  tStep 2Then you immediately get  I  =  −      1          c        ∫  d  t  =  −      1          c            a    r    s    i    n    h    (      c        e          −      x        )  +  c  o  n  s  t  ..For   c  &amp;lt;  0 there&#039;s a similar substitution,       −    c        e          −      x        =  sin  ⁡  t

Antiderivatives such as ∫ d x </mrow> <msqrt>

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