Incorrect answer in indefinite integration (HP Prime)

[Thomas Klemm]

(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:

\(
\begin{align}
F(x)
&= \int_{-2}^{x} \log(t+2) \; \mathrm{d}t \\
\\
&= (t+2) \log(t+2) - t \Big|_{-2}^x \\
\\
&= (x+2) \log(x+2) - x - 2 \\
\end{align}
\)