How would I go about finding the roots of (ex−1)−karctan⁡(x)=0?

If k>0
Let
${\displaystyle f\left(x\right)=e^x-1-k{\tan }^{-1}x\Rightarrow f\prime \left(x\right)=e^x-\frac{k}{1+x^2},f\prime \prime \left(x\right)=e^x+\frac{k}{{(1+x^2)}^2}>0}$
${\displaystyle \Rightarrow }$ f′(x) is an increasing function ${\displaystyle \Rightarrow f\prime \left(x\right)>f\prime (-\mathrm{\infty })>0\Rightarrow f\left(x\right)}$ is an increasing function. Hence f(x)=0, will have at most one real root. As ${\displaystyle f(-\mathrm{\infty })=-1+\frac{k\pi }{2}}$ and ${\displaystyle f\left(\mathrm{\infty }\right)>0}$, for one real root by IVT ${\displaystyle f(-\mathrm{\infty })<0}$. Finally this gives ${\displaystyle 0<k<2\pi }$. The one root is x=0.
For ${\displaystyle k>\frac{2}{\pi }}$, both ${\displaystyle f(-\mathrm{\infty })>}$ and ${\displaystyle f\left(\mathrm{\infty }\right)>0}$ so f(x)=0 will have 0,2,4,... number of real roots. Since x=0 is essentially a root. So there will two real roots if ${\displaystyle k>\frac{2}{\pi }}$. Further, since f′′(x)>0, for the function f(x) can have atmost one min, this rules out more than two real roots.

Of course x=0 is always a solution. Otherwise, write the equation as $k=\frac{e^x-1}{\arctan (x)}$ Call the right side f(x). The singularity at x=0 is removable, with $\underset{x\to 0}{lim}f(x)=1$. We also have $\underset{x\to -\mathrm{\infty }}{lim}f(x)=\frac{2}{\pi }$ and $\underset{x\to +\mathrm{\infty }}{lim}f(x)=+\mathrm{\infty }$ It appears that f(x) is increasing. If so, for $\frac{2}{\pi }<k<1$ and $1<k<\mathrm{\infty }$ there are two real roots (x=0 and the root of f(x)=k), otherwise there is only x=0.

Let $f(x)=e^x-1-k\arctan (x)$. We have f(0)=0. If k=0, obviously x=0 is the only solution. If $k\ne 0$, then $f\prime (x)=e^x-\frac{k}{1+x^2}$, so that $f\prime (x)=0\iff (1+x^2)e^x=k$. Since $(1+x^2)e^x$ is increasing, goes to 0 at $-\mathrm{\infty }$ and goes to $\mathrm{\infty }\text{ at }\mathrm{\infty }$ we see that f′ has exactly one real zero, say f′(x0)=0. We see that f decreases on $(-\mathrm{\infty },x_0)$ and increases on $(x_0,\mathrm{\infty })$, so that f attains its minimum at $x_0$ Since we know that f(0)=0, it must be that $x_0<0$ and x=0 is the only solution.