Plotting $2 \arctan x + \arctan \frac{2x}{x^2-1}$ I need help to plot the graph $$y = 2 \arctan x + \arctan \frac{2x}{x^2-1}.$$ I'm trying to simplify the expression by setting
$$
\tan(y) = \tan\left(2 \arctan x + \arctan \frac{2x}{x^2-1}\right).
$$
However, I'm always ending up with the RHS being equal to $0$ which gives me $y = 0$.
Plotting the graph shows me that it consists of three regions, one of which the y - value is $-\pi$, one where it's 0, and one where it's $\pi$. How can you from the given function determine these regions? Is there anything special you have to do before starting to simplify the expression. I assume it has to do something with the $x^2-1$ that in some way or another changes the sign of the expression for some given x's and thus, changing the overall value of the function of y.
Thanks for the help!
 A: Write $f(x) = 2 \arctan x + \arctan \frac{2x}{x^2 - 1}$.
Your algebra is correct that $\tan f(x) = 0$ whenever $f(x)$ is defined.
However, there are many values of $\theta$ such that $\tan \theta = 0$. In fact, $\tan \theta = 0$ if and only if $\theta$ is an integer multiple of $\pi$.
So we know that for all $x$ such that $|x| \neq 1$, we have $f(x)$ is a multiple of $\pi$. In other words, $f(x) \in \pi \mathbb{Z}$, where $\pi \mathbb{Z}$ is the set $\{\pi n \mid n \in \mathbb{Z}\}$.
Now the domain of $f$ is $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$. We see that $f$ is continuous on its domain.
Let $I_1 = (-\infty, -1)$, $I_1 = (-1, 1)$, and $I_3 = (1, \infty)$.
We see that each $I_j$ is connected. Therefore, the image $f(I_j)$ is connected and non-empty. Thus, $f : I_j \to \pi \mathbb{Z}$ must be constant, since $\pi \mathbb{Z}$ has the discrete topology and hence the only connected subsets are those with at most one element.
Now, we simply compute $f$ at some convenient values.
We see that $\lim\limits_{x \to -\infty} f(x) = -\pi$, hence $f(x) = -\pi$ for $x \in I_1$.
We see that $f(0) = 0$, hence $f(x) = 0$ for all $x \in I_2$.
And we se that $\lim\limits_{x \to \infty} f(x) = \pi$, hence $f(x) = \pi$ for all $x \in I_3$.
