If $\int e^{x + \tan^{-1}x}\left(\dfrac{x^2 + 2}{\sec^2(\tan^{-1}x)}\right)dx = e^{f(x)} + c$, then $\ldots$

The given integral is $I = \int e^{x + \tan^{-1}x}\left(\dfrac{x^2 + 2}{\sec^2(\tan^{-1}x)}\right)dx$.

Using the identity $\sec^2\theta = 1 + \tan^2\theta$, we have $\sec^2(\tan^{-1}x) = 1 + \tan^2(\tan^{-1}x) = 1 + x^2$.

Substituting this into the integral gives:

$I = \int e^{x + \tan^{-1}x}\left(\dfrac{x^2 + 2}{1 + x^2}\right)dx$

$I = \int e^{x + \tan^{-1}x}\left(\dfrac{x^2 + 1 + 1}{1 + x^2}\right)dx$

$I = \int e^{x + \tan^{-1}x}\left(1 + \dfrac{1}{1 + x^2}\right)dx$

Let $t = x + \tan^{-1}x$. Differentiating with respect to $x$, we get:

$dt = \left(1 + \dfrac{1}{1 + x^2}\right)dx$

The integral becomes:

$I = \int e^t dt = e^t + c$

Substituting back $t = x + \tan^{-1}x$, we get:

$I = e^{x + \tan^{-1}x} + c$

Comparing this with the given expression $e^{f(x)} + c$, we find:

$f(x) = x + \tan^{-1}x$

To determine the monotonicity of $f(x)$, we find its derivative:

$f'(x) = 1 + \dfrac{1}{1 + x^2}$

Since $x^2 \ge 0$ for all real $x$, $1 + x^2 \ge 1$, which implies $\dfrac{1}{1 + x^2} > 0$.

Thus, $f'(x) > 1 > 0$ for all $x \in R$.

Therefore, $f(x)$ is strictly increasing on $R$.

Answer: $f(x)$ is strictly increasing on $R$.