When are absolute values needed in the antiderivative?I just tackled a STEP I past paper question (Q4(iii), 2004) which asked me to determine the antiderivative for this integral: ∫ 1 ( x + 2 ) x 2 + 4 x − 5 d xwhich (after substituting x + 2 3 = sec ⁡ ( t )) gives: 1 3 sec − 1 ⁡ ( x + 2 3 ) + CThis is also the answer given by Cambridge in the answer booklet. However, I also put this integral into an online integral calculator and one of the possible antiderivatives was − 1 3 arcsin ⁡ ( 3 ∣ x + 2 ∣ ) + C 1 This is (according to Desmos) is equivalent to: 1 3 sec − 1 ⁡ ( ∣ x + 2 ∣ 3 ) + C 2 which is not at all the same function as the one I got. However, looking at the graphs (of the original functions and the antiderivatives), it would seem that the integral calculator is correct. I am presuming that Cambridge simplified the answer for STEP.I can see that the quadratic in the radical has a minimum point at x = − 2, but we haven't really been taught much about absolute values and discontinuities, except for ∫ 1 x d x = ln ⁡ ∣ x ∣ + C,which we were told was because ln(x) can't take negative values. Yet, arcsin(x) above can, and there's an absolute value there.Could someone explain to me how all of this works, please?

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When are absolute values needed in the antiderivative?I just tackled a STEP I past paper question (Q4(iii), 2004) which asked me to determine the antiderivative for this integral:  ∫      1          (      x      +      2      )                        x          2                +        4        x        −        5              d  xwhich (after substituting             x      +      2        3    =  sec  ⁡  (  t  )) gives:      1    3        sec          −      1        ⁡      (                  x        +        2            3        )    +  CThis is also the answer given by Cambridge in the answer booklet. However, I also put this integral into an online integral calculator and one of the possible antiderivatives was  −      1    3    arcsin  ⁡      (          3              ∣        x        +        2        ∣              )    +      C    1  This is (according to Desmos) is equivalent to:      1    3        sec          −      1        ⁡      (                  ∣        x        +        2        ∣            3        )    +      C    2  which is not at all the same function as the one I got. However, looking at the graphs (of the original functions and the antiderivatives), it would seem that the integral calculator is correct. I am presuming that Cambridge simplified the answer for STEP.I can see that the quadratic in the radical has a minimum point at   x  =  −  2, but we haven&#039;t really been taught much about absolute values and discontinuities, except for  ∫      1    x    d  x  =  ln  ⁡      ∣    x    ∣    +  C,which we were told was because ln(x) can&#039;t take negative values. Yet, arcsin(x) above can, and there&#039;s an absolute value there.Could someone explain to me how all of this works, please?

Eleanor Luna · Accepted Answer

Step 1You can learn a lot by really understanding the single antiderivative   ∫      1    t    d  t  ..We are tempted to interpret this as                     (1)                              ∫          1          x                          1          t                d        t        =        ln        ⁡        x        ,                (        x        &amp;gt;        0        )            but we know that the choice of the base point 1 is arbitrary. We could find another antiderivative by looking at       ∫    4    x        1    t    d  t  =  ln  ⁡  x  −  ln  ⁡  4.Step 2This ambiguity leads to the &quot;up to an additive constant C&quot;, as both of these are acceptable antiderivatives when   x  &amp;gt;  0.But what about when   x  &amp;lt;  0? Then       ∫    1    x        1    t    d  t doesn&#039;t converge - the domain of integration includes the point 0, near which 1/t  shoots off to infinity. On the other hand, if   x  &amp;lt;  0, then                     (2)                              ∫                      x                                −            1                                    1          t                d        t        =                  ∫                      1                                −            x                                    1          t                d        t        =        ln        ⁡        (        −        x        )        .                (        x        &amp;lt;        0        )            (The equality comes from substituting   t  ↦  −  t and then writing       ∫    a    b    =  −      ∫    b    a  .) Of course, we know this is really only true up to a constant C.We know that (1) applies for all   x  &amp;gt;  0 and (2) applies for all   x  &amp;lt;  0. There is no antiderivative corresponding to   x  =  0, since every integral of the form      ∫    A    0        1    t    d  tdiverges. Thus a complete story would be to say that the antiderivative of       1    t   is  ∫      1    t    d  t  =      {                            ln          ⁡          x          +          C                          if           x          &amp;gt;          0          ,                                      ln          ⁡          (          −          x          )          +          C                          if           x          &amp;lt;          0.                        It just happens to be that this is given by   ln  ⁡  |  x  |  +  C that&#039;s sort of a coincidence.

Jeffery Clements · Accepted Answer

Step 1Your antiderivative is wrong when   x  &amp;lt;  −  1 (its derivative has the wrong sign).If you use the substitution   (  x  +  2  )      /    3  =  sec  ⁡  (  t  ) (I don&#039;t think you really used   x  =  sec  ⁡  (  t  )), your integral becomes  ∫            tan      ⁡      (      t      )              3                        sec          2                ⁡        (        t        )        −        1                d  tNow       sec    2    ⁡  (  t  )  −  1  =      tan    2    ⁡  (  t  ), so it may be tempting to write this as   ∫  d  t      /    3  =  t      /    3  +  C leading to your answer. But this is wrong: if   tan  ⁡  (  t  )  &amp;lt;  0 (which corresponds to   x  &amp;lt;  −  5) we will have             sec      2        ⁡    (    t    )    −    1    =  −  tan  ⁡  (  t  ). So a correct answer would be       {                                        1            3                                sec                          −              1                                ⁡                      (                                          x                +                2                            3                        )                    +          C                                       if                           x              &amp;gt;              1                                                            −                      1            3                                sec                          −              1                                ⁡                      (                                          x                +                2                            3                        )                    +          C                                       if                           x              &amp;lt;              −              5                                              which (with different C in the second case, but there&#039;s no need for the C&#039;s to be the same) could be written as      1    3        sec          −      1        ⁡      (                            |                x        +        2                  |                    3        )    +  COf course, if for some reason you&#039;re only interested in   x  &amp;gt;  5 your answer would be OK.

When are absolute values needed in the antiderivative? I just tackled a STEP I past paper question

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New Questions in Calculus 1