3
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Aug 27, 2021 at 5:04
  • 3
    $\begingroup$ Both the answers differ by a 'constant' term only. Nothing is wrong in that. $\endgroup$ Commented Aug 27, 2021 at 5:05
  • 1
    $\begingroup$ There must be a duplicate with zillions of linked questions... $\endgroup$ Commented Aug 31, 2021 at 16:13
  • $\begingroup$ @metamorphy there's an faq written by Xander sometimes ago here $\endgroup$ Commented Sep 3, 2021 at 7:30

1 Answer 1

Reset to default
4
$\begingroup$

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.

$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.