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In looking at the equality

$$\int \frac{a}{b(c-x)}dx = \int dt$$

I obtained different answers via different methods.

Via one method, I got

$$- \frac{a}{b} \ln(c-x) = t+C$$

Via another, I got

$$- \frac a b \ln(b(c-x)) = t+C$$

How is this possible?

PrincessEev
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    A couple things... The best place for question content is in the question body, not at an external location which requires a click; this is usually best accomplished with MathJAX. Also, there is no question here... What type of answer are you expecting? – abiessu Aug 27 '21 at 05:04
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    Both the answers differ by a 'constant' term only. Nothing is wrong in that. – Ilovemath Aug 27 '21 at 05:05
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    There must be a duplicate with zillions of linked questions... – metamorphy Aug 31 '21 at 16:13
  • @metamorphy there's an faq written by Xander sometimes ago here – Arctic Char Sep 03 '21 at 07:30

1 Answers1

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The $C$ in both answers are not the same, assuming $a,b,c$ are constants.

Note that

$$-\frac a b \ln(b(c-x)) = - \frac{a}{b} \ln(b) - \frac a b \ln(c-x) = t + K$$

If we set $K = C - a \ln(b) / b$, then it becomes equivalent to your first result.

PrincessEev
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