Using this rule: Integral Calculus "Rational Function Rule" ∫ 1 a 2 − x 2 d x = 1 2 a log e ⁡ ( a + x a − x ) + cI have integrated a function, but my question is if I'm solving for x, in the next line I must raise both sides to the base of e, and there is still + c (constant) on the end of the integrated function, so do I leave it as plus c on the RHS on the equation (not do anything to it, so when I want to work it out by substituting ( x 1 , y 1 ) at the end it will be x 1 = … y 1 … + c) or when I raise both sides to the base of e does c, the constant become c ∗ e … instead of + c?Any help would be great.

Question

Using this rule: Integral Calculus &quot;Rational Function Rule&quot;  ∫      1                  a                  2                    −              x                  2                        d  x  =      1          2      a            log          e        ⁡      (                  a        +        x                    a        −        x              )    +  cI have integrated a function, but my question is if I&#039;m solving for   x, in the next line I must raise both sides to the base of   e, and there is still   +  c (constant) on the end of the integrated function, so do I leave it as plus   c on the RHS on the equation (not do anything to it, so when I want to work it out by substituting   (      x    1    ,      y    1    ) at the end it will be       x    1    =  …      y    1    …  +  c) or when I raise both sides to the base of   e does   c, the constant become   c  ∗  e  … instead of   +  c?Any help would be great.

seraphinod · Accepted Answer

The antiderivative exists in each of the follwing intervals :  (  −  ∞  ,  −  a  )  ;  (  −  a  ,  a  )  ;  (  a  ,  +  ∞  )the result should be      1    2    ln  ⁡  (                    a        +        x                    a        −        x              λ    )  =      1    2    (  ln  ⁡  (      |              a      +      x              a      −      x            |    )  −  ln  ⁡  (      |    λ      |    )  =      1    2    ln  ⁡  (      |              a      +      x              a      −      x            |    )  +  C

hawatajwizp · Accepted Answer

If   y  =  x  +  C, then       e          y        =      e          x      +      C        =      e    x        e          C      , so the additive arbitrary constant becomes a multiplicative arbitrary constant.

Using this rule: Integral Calculus "Rational Function Rule" ∫ 1

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