Evaluating the integral $$ \int \frac{x^2 -2x+13}{(x-2)(x^2+2x+5)} dx $$ using partial fractions I obtained $$ \ln \vert x -2|-2 \arctan\left(\frac{x+1}{2}\right) +c_1 $$ The answer given in the textbook is $$ \ln \vert x -2|+2 \arctan\left(\frac{2}{x+1}\right) +c_2$$ WolframAlpha gave the answer $$ \ln (2-x) +2 \arctan\left(\frac{2}{x+1}\right) +c_3$$ when inputting the initial fraction directly, and the answer $$ \ln (x -2)-2 \arctan\left(\frac{x+1}{2}\right) +c_4 $$ when I entered it in partial fraction form.
I tried plotting the graphs of each of these functions (shown below) to understand what's going on, but I'm still confused as to which of these is/are correct, and where the discrepancies arise from.
In each of the regions $(- \infty,1) $,$(-1,2) $ and $(2, + \infty) $ each graph is the same shape (if it is defined in that region), so the difference is only in the constants of integration. However, the WolframAlpha answer defined for $x<2$ (which I have taken to be correct for this region) is not continuous in this region, whereas my answer is, which I presume makes it incorrect?
The textbook answer follows the shape of the WolframAlpha answer for $x<2$, but is also defined for $x>2$. For it to follow the shape of both of the WolframAlhpa graphs simultaneously, it requires them to have different constants of integration; if both of the WolframAlpha answers are required to fully describe this integral in two regions, can they have different constants of integration?
The conversion of $\ln|x-2|$ to $\ln(2-x)$ is a product of forgetting the antiderivatives of $\frac1x$ are of the form $\ln x+c_+$ for $x>0$ and $\ln(-x)+c_-$ for $x<0$. (As for its being $2-x$ rather than $x-2$, I'm guessing that either your partial fractions included a $2-x$ denominator instead of an $x-2$ one, or their algorithm prefers the $a-x$ option over $x-a$.)
The identity $\arctan\frac1x=\frac{\pi}{2}-\arctan x$ for $x>0$ (for $x<0$, the fraction becomes $\frac{-\pi}{2}$) explaines the $+2\arctan\frac{2}{x+1}$ answers.
Let's discuss this more carefully than the software did. The odd antiderivative of $\frac{1}{1+x^2}$ is $\arctan x$, and unlike the case of $1/x$, there's no discontinuity to complicate the family of antiderivatives. You're trying to integrate $\frac{1}{x-2}-\frac{4}{x^2+2x+5}$. So if we want an $\Bbb R\mapsto\Bbb R$ antiderivative, the most general result is $\ln|x-2|-2\arctan\frac{x+1}{2}+C$, where $C$ is locally constant but can assume different values on either side of $x=2$. (This is due to integrating the $\frac{1}{x-2}$.) It's erroneous of software to try $2\arctan\frac{2}{x+1}$, since unlike any true antiderivative of $\frac{1}{x^2+4x+5}$ this function has different left- and right-hand limits at $x=-1$.