Antiderivatives of 1 x − 1 I was told to find the anti-derivative of this problem: ∫ 2 x 2 − 13 x + 18 x − 1 d xI solved the problem - first dividing and then finding the anti-derivative: x 2 − 11 x + 7 ( ln ⁡ ( x − 1 ) ) + cThe given answer was the same as mine, but they put the x − 1 in ln ⁡ ( x − 1 ) in absolute value ln ⁡ | x − 1 | . Why?

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Antiderivatives of       1          x      −      1      I was told to find the anti-derivative of this problem:  ∫            2              x        2            −      13      x      +      18              x      −      1        d  xI solved the problem - first dividing and then finding the anti-derivative:      x    2    −  11  x  +  7  (  ln  ⁡  (  x  −  1  )  )  +  cThe given answer was the same as mine, but they put the   x  −  1 in   ln  ⁡  (  x  −  1  ) in absolute value   ln  ⁡      |    x  −  1      |  . Why?

odmeravan5c · Accepted Answer

Expanation:Consider what happens if   x  &amp;lt;  1: The original function makes complete sense, so there ought to be an antiderivative, but your expression doesn&#039;t make sense. Calculate it, and   ln  ⁡  (  1  −  x  ) appears. The two antiderivatives for   x  &amp;lt;  1 and   x  &amp;gt;  1 are most easily stitched together with an absolute value.

Finley Mckinney · Accepted Answer

Step 1A lot of people are telling you it&#039;s a domain error, but I will try to give an explanation for a beginner level calculus student looking at this problem. This isn&#039;t exactly your problem, but explains why we include the absolute value around ln|x|.Look at this integral       ∫          −      2              −      1                  1      x        d  xNow if you graph this function it is very clear that this function is defined on this interval, so you should be able to take the area under the curve right?Step 2Since you know that the antiderivative of       1    x   is ln(x), the result should be   ln  ⁡  (  x  )            |              −      2              −      1        =  ln  ⁡  (  −  1  )  −  ln  ⁡  (  −  2  )ln(x) is not defined for   x  ≤  0 (well unless we&#039;re in complex plane, but thats a whole other can of worms)! So in order to make sure this integral is correctly evaluated for   x  &amp;lt;  0, we need to place absolute value signs around the argument       ∫          −      2              −      1                  1      x        d  x  =  ln  ⁡      |    −  1      |    −  ln  ⁡      |    −  2      |    =  −  ln  ⁡  2

Antiderivatives of 1 x − 1 </mrow> </mfrac>

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