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integral(ln(x+2)dx

why is there a -2 at the end?
(02-28-2024 07:00 PM)ReinXXL Wrote: [ -> ]integral(ln(x+2)dx

why is there a -2 at the end?

Why not? Indefinite integrals are always +/- any arbitrary constant value, of which "-2" is a special case. I assume this results from Xcas implementation.
I had the same thought. Was wondering how/why XCAS came up with a constant equaling 2 as opposed to something else!
Maybe it is the airspeed velocity of an unladen African sparrow?
(02-29-2024 02:46 PM)KeithB Wrote: [ -> ]Maybe it is the airspeed velocity of an unladen African sparrow?

And because the airspeed velocity is negative (-2), perhaps it is flying backwards :)
Why -2?

You can rewrite the answer x*ln(x+2)-x+2*ln(x+2)-2 as (x+2)*ln(x+2)-(x+2).
(02-28-2024 07:00 PM)ReinXXL Wrote: [ -> ]why is there a -2 at the end?

We can consider the singularity at $$x=-2$$ a natural lower bound of the definite integral.

This choice of the integral constant makes it $$0$$ at that value:

\begin{align} F(x) &= \int_{-2}^{x} \log(t+2) \; \mathrm{d}t \\ \\ &= (t+2) \log(t+2) - t \Big|_{-2}^x \\ \\ &= (x+2) \log(x+2) - x - 2 \\ \end{align}
(02-29-2024 03:42 PM)carey Wrote: [ -> ]
(02-29-2024 02:46 PM)KeithB Wrote: [ -> ]Maybe it is the airspeed velocity of an unladen African sparrow?

And because the airspeed velocity is negative (-2), perhaps it is flying backwards

Or flying West?
Not really mysterious, it's a linear change of variable.
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