Came from a homework question but two different but same answers.
two approach, but reach equivalent answers fine.
Checked with differentiation,
Fine
However, the graphs look like reflections of each other?
Shouldn't the gradient be reflections??
That book is all too familiar lol. Okay so I've played around with $\frac{1}{sinh(x)}$ and feel ready to answer this question.
The way you asked your question was low key confusing, so I may discuss more than what is needed but nevertheless will hopefully provide a good reason for why those two, suitable antiderivatives are reflections of each other.
Let's look at the graph of $\frac{1}{sinh(x)}\space\space(=cosech(x))$
Click for graph of: $$\frac{1}{sinh(x)}$$ As you can see, there is a vertical asymptote at $x=0$, this asymptote is the reason for those two antiderivatives being reflections of each other. I'll explain: $$-2coth^{-1}(e^x)+c$$ $$-2tanh^{-1}(e^x)+c$$ are two suitable antiderivatives as you know.
$-2coth^{-1}(e^x)$ corresponds to the area under the $x>0$ branch of $cosech(x)$
$-2tanh^{-1}(e^x)$ corresponds to the area under the $x<0$ branch of $cosech(x)$
You can see this from the plot you made of the two functions.
Namely if you were evaluating $$\int_1^5 \frac{1}{sinh(x)} dx $$, it would be easiest to evaluate it as: $$[-2coth^{-1}(e^x)]_1^5$$ You could try use $-2tanh^{-1}(e^x)$ but you would quickly run into the problem of $ln(\text{negative number})$ and for further math, the notion of $ln(\text{negative number})$ is not defined yet.
So why is this? it all comes down to the fact that for all real numbers $x$, $$e^x>0$$
Compare the plot of $-2coth^{-1}(x)$ and $-2coth^{-1}(e^x)$
As you can see, since $e^x$ is always positive, the $x<0$ part of $-2coth^{-1}(x)$ is absent.
Try the same analysis on $-2tanh^{-1}(e^x)$, if it is not making complete sense
Now with that foundation in mind, onto your answering your question:
A function isn't just defined by the expression,$f(x)=...$, but also its domain.
Namely, if $\frac{df}{dx}=h(x)$ and $\frac{dg}{dx}=h(x)$ This does not imply $f(x)$ is the same as $g(x)$
Both $-2coth^{-1}(e^x)+c$ and $-2tanh^{-1}(e^x)+c$ have the same derivative: $$\frac{1}{sinh(x)}$$ However both functions exist for different domains (one has $x>0$ and the other $x<0$).
The reason they are reflections of each other, is because the graph of $$\frac{1}{sinh(x)}$$ the negative portion of the graph has an area equal in magnitude and opposite in sign to the area under the positive portion of the graph (it is symmetric about the "$y=-x$" line). Look at the graph to convince yourself of this.
Conclusion: Hence both are valid and represent different portions of the $cosech(x)$ graph. They are reflections due to the symmetries of $cosech(x)$. Don't worry too much about this for further math though.
A better antiderivative for $\frac{1}{sinh(x)}$ is: $$-\ln\left(\left|\operatorname{csch}\left(x\right)+\coth\left(x\right)\right|\right)+c $$ Plot a graph of this to see why. However this result is not easily obtained by hand. (I used to integral calculator to get it).
This whole question highlights the importance of: $$\int \frac{1}{x} dx = ln|x|+c \space\space\space\text{ not } =ln(x)+c$$
Hope this helps and really, it is just extra stuff. Best of luck with your studies!