Clarification about the Antiderivative of x − 1 / 3 .In some textbook the antiderivative of x − 1 / 3 is written as ∫ x − 1 / 3 d x = 3 2 x 2 / 3 + C ,where C is a constant. But should not the following function also be considered as an antiderivative of x − 1 / 3 ? F ( x ) = { 3 2 x 2 / 3 + C 0 , if x &gt;0 3 2 x 2 / 3 + C 1 , if x &lt;0 When C 0 ≠ C 1 , F(x) cannot be written as 3 2 x 2 / 3 + C.Edit: I would like to clarify: which one of the following should be the correct answer to ∫ x − 1 / 3 d x:- 3 2 x 2 / 3 + C- or { 3 2 x 2 / 3 + C 0 , if x &gt;0 3 2 x 2 / 3 + C 1 , if x &lt;0 ?

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Clarification about the Antiderivative of       x          −      1              /            3      .In some textbook the antiderivative of       x          −      1              /            3       is written as  ∫      x          −      1              /            3            d    x  =      3    2        x          2              /            3        +  C  ,where C is a constant. But should not the following function also be considered as an antiderivative of       x          −      1              /            3      ?  F  (  x  )  =      {                                        3            2                                x                          2                              /                            3                                +                      C            0                    ,                       if                           x                        &amp;gt;0                                                            3            2                                x                          2                              /                            3                                +                      C            1                    ,                       if                           x                        &amp;lt;0                                  When       C    0    ≠      C    1  , F(x) cannot be written as       3    2        x          2              /            3        +  C.Edit: I would like to clarify: which one of the following should be the correct answer to   ∫      x          −      1              /            3            d    x:-       3    2        x          2              /            3        +  C- or       {                                        3            2                                x                          2                              /                            3                                +                      C            0                    ,                       if                           x                        &amp;gt;0                                                            3            2                                x                          2                              /                            3                                +                      C            1                    ,                       if                           x                        &amp;lt;0                                    ?

yen1291kp6 · Accepted Answer

Step 1The integrand       x          −      1              /            3       has a singularity at   x  =  0 and you cannot integrate across it. So the expressions in the negatives and in the positives are independent of each other and you can very well consider two constants. The derivative of the antiderivative and the original function do match wherever the function is defined.Step 2In practice it is often the case that you integrate in a single interval, so that the existence of another branch doesn&#039;t matter and writing C (which disappears in the definite integral) is enough.This generalizes to functions with more singularities, say  2  ∫            d      x                      x        2            −      1        =      {                            x          &amp;lt;          −          1                          →          log          ⁡                      |                                          1                −                x                                            1                +                x                                      |                    +                      C            0                    ,                                      −          1          &amp;lt;          x          &amp;lt;          1                          →          log          ⁡                      |                                          1                −                x                                            1                +                x                                      |                    +                      C            1                    ,                                      x          &amp;gt;          1                          →          log          ⁡                      |                                          1                −                x                                            1                +                x                                      |                    +                      C            2                    .

Clarification about the Antiderivative of x^{-1/3}

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