Integration - involving logarithmI have a slight confusion on how to integrate functions of the form: ∫ a x d xSuppose we have the following function: ∫ − 2 x d xThere are two ways we can proceed to integrate this function. One is to treat the − 2 sign as a constant and take it out of the integration function: − 2 ∫ 1 x = − 2 ln ⁡ x = ln ⁡ 1 x 2 Another way to do this is to treat the − 2 as part of the integration variable: ∫ 1 − 0.5 x d x = − 2 ln ⁡ ( − 0.5 x ) This seems to obtain two different answers. Which method is correct? Or is there something that I'm missing?

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Integration - involving logarithmI have a slight confusion on how to integrate functions of the form:  ∫      a    x    d  xSuppose we have the following function:  ∫            −      2        x    d  xThere are two ways we can proceed to integrate this function. One is to treat the   −  2 sign as a constant and take it out of the integration function:  −  2  ∫      1    x    =  −  2  ln  ⁡      x    =  ln  ⁡            1              x        2            Another way to do this is to treat the   −  2 as part of the integration variable:  ∫      1          −      0.5      x        d  x  =  −  2  ln  ⁡      (    −    0.5    x    )  This seems to obtain two different answers. Which method is correct? Or is there something that I&#039;m missing?

Mohammad Cantrell · Accepted Answer

To be proper, you should use that  ∫      1    x    d  x  =  ln  ⁡      |    x      |    +  Cnot just   ln  ⁡  x (as defining the logarithm for negative numbers requires some care). Given this, you found two antiderivatives:  ln  ⁡      1          x      2           and   −  2  ln  ⁡  (  .5      |    x      |    )The second one can be written as  −  2  ln  ⁡  (  .5  )  −  2  ln  ⁡      |    x      |    =  −  2  ln  ⁡  (  .5  )  +  ln  ⁡      1          x      2      These differ by a constant, which is not a problem.

Kayla Mcdowell · Accepted Answer

Both methods are correct. The results differ by a constant:  −  2  ln  ⁡      |        −    0.5    ⋅    x        |    =  −  2  ln  ⁡      |    x      |    −  2  ln  ⁡  (  0.5  )  =  −  2  ln  ⁡      |    x      |    +  CKeep in mind that  ∫      1    x      d  x  =  ln  ⁡      |    x      |    +  C

Integration - involving logarithm I have a slight confusion on how to integrate functions of the form: int a/x dx

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