All possible antiderivatives for Integration by PartsJust had a quick inquiry with regards to the formula for Integration by Parts. If I'm not mistaken, the formula states that ∫ f ′ ( x ) g ( x ) = f ( x ) g ( x ) − ∫ f ( x ) g ′ ( x )However, in the case that I try to substitute an antiderivative with a valid constant, the formula does not appear to work. I attempted to use a simple product such as 7 x ( x 2 ), using 7x as f′(x) and x 2 as g(x) respectively. I found that using 7 2 x 2 + 5 for the antidervative for 7x does not work in the formula, while 7 2 x 2 without the constant does indeed work. I am sure that I am missing something, however, why do both of these solutions not work, even though both are valid antiderivatives?

Question

All possible antiderivatives for Integration by PartsJust had a quick inquiry with regards to the formula for Integration by Parts. If I&#039;m not mistaken, the formula states that   ∫      f    ′    (  x  )  g  (  x  )  =  f  (  x  )  g  (  x  )  −  ∫  f  (  x  )      g    ′    (  x  )However, in the case that I try to substitute an antiderivative with a valid constant, the formula does not appear to work. I attempted to use a simple product such as   7  x  (      x    2    ), using 7x as f′(x) and       x    2   as g(x) respectively. I found that using       7    2        x    2    +  5 for the antidervative for 7x does not work in the formula, while       7    2        x    2   without the constant does indeed work. I am sure that I am missing something, however, why do both of these solutions not work, even though both are valid antiderivatives?

pyphekam · Accepted Answer

Step 1Say you pick   f  (  x  )  =      7    2        x    2    +  5,   g  (  x  )  =      x    2  , and are doing   ∫      f    ′    (  x  )  g  (  x  )    d  x  =  ∫  (  7  x  )      x    2      d  x  =  ∫  7      x    3      d  x  =      7    4        x    4    +  C  ..Now, if you try using integration by parts with the anti-derivative you pick, you have                    ∫                  f          ′                (        x        )        g        (        x        )                d        x                            =        f        (        x        )        g        (        x        )        −        ∫                  g          ′                (        x        )        f        (        x        )                d        x                                          =                  (                      7            2                                x            2                    +          5          )                (                  x          2                )        −        ∫        2        x                  (                      7            2                                x            2                    +          5          )                        d        x                                          =                  7          2                          x          4                +        5                  x          2                −        ∫        (        7                  x          3                +        10        x        )                d        x                                          =                  7          2                          x          4                +        5                  x          2                −                  (                      7            4                                x            4                    +          5                      x            2                    +          D          )                                                  =                  7          2                          x          4                −                  7          4                          x          4                +        5                  x          2                −        5                  x          2                −        D                                          =                  7          4                          x          4                −        D        .            That is, the same answer, up to a constant.Step 2So long as you use the same antiderivative in both instances of f(x) on the right hand side, it will work out. Recall that if f(x) is one antiderivative, then every antiderivative is of the form   f  (  x  )  +  D, with D a constant. So you would get:                    (        f        (        x        )        +        D        )        g        (        x        )                            −        ∫        (        f        (        x        )        +        D        )                  g          ′                (        x        )                d        x                                          =        f        (        x        )        g        (        x        )        +        D        g        (        x        )        −        ∫        f        (        x        )                  g          ′                (        x        )                d        x        −        ∫        D                  g          ′                (        x        )                d        x                                          =        f        (        x        )        g        (        x        )        +        D        g        (        x        )        −        ∫        f        (        x        )                  g          ′                (        x        )                d        x        −        D        ∫                  g          ′                (        x        )                d        x                                          =        f        (        x        )        g        (        x        )        +        D        g        (        x        )        −        ∫        f        (        x        )                  g          ′                (        x        )                d        x        −        D        (        g        (        x        )        +        E        )                                          =        f        (        x        )        g        (        x        )        +        D        g        (        x        )        −        D        g        (        x        )        −        ∫        f        (        x        )                  g          ′                (        x        )                d        x        −        D        E                                          =        f        (        x        )        g        (        x        )        −        ∫        f        (        x        )                  g          ′                (        x        )                d        x            (because that final constant gets &quot;absorbed&quot; into the indefinite integral).

All possible antiderivatives for Integration by Parts Just had a quick inquiry with regards to the

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