Is $\arctan(t)$ an energy or power signal?
The typical inverse tangent function maps the input range of $ t \in (-\infty,\infty)$ into an output range of $(-\pi/2, \pi/2)$ as in the figure below:



Based on this, its values are bounded for all $t$. Yet, since the intergal of its square is unbounded, then it cannot be an energy signal;i.e.,

$$ \int_{-\infty}^{\infty} |\tan^{-1}(t)|^2 dt ~~~~~\to \infty $$

Moreover the average power of this signal is nonzero and finite, hence this is a power signal;i.e.,

$$ 0 &lt; \lim_{T \to \infty} \frac{1}{T}\int_{-T/2}^{T/2} |\tan^{-1}(t)|^2 dt ~~~~~ &lt; \infty $$

To evaluate the last integral, use the following MATLAB figure to see that

:

the square of the inverse tangent is always less than $(\pi/2)^2$ for all $t$. Hence to find an upper bound on the integral we can use $|\tan^{-1}(t)|^2 &lt; (\pi/2)^2 = K$. Then the integral becomes

$$ 0 &lt; \lim_{T \to \infty} \frac{1}{T}\int_{-T/2}^{T/2} |\tan^{-1}(t)|^2 dt &lt; \lim_{T \to \infty} \frac{1}{T}\int_{-T/2}^{T/2} K dt = \lim_{T \to \infty} \frac{1}{T} K \cdot T = K ~~~~~ &lt; \infty $$