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Let $I = \int \frac{e^{m \tan^{-1}x}}{(1 + x^2)^2} dx$.

Let $\tan^{-1}x = \theta \implies x = \tan\theta$.
Differentiating with respect to $\theta$, we get $dx = \sec^2\theta \, d\theta$.

Substituting these into the integral:
$I = \int \frac{e^{m\theta}}{(1 + \tan^2\theta)^2} \sec^2\theta \, d\theta$
$I = \int \frac{e^{m\theta}}{\sec^4\theta} \sec^2\theta \, d\theta = \int e^{m\theta} \cos^2\theta \, d\theta$

Using the identity $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$:
$I = \frac{1}{2} \int e^{m\theta} (1 + \cos 2\theta) \, d\theta = \frac{1}{2} \int e^{m\theta} d\theta + \frac{1}{2} \int e^{m\theta} \cos 2\theta \, d\theta$

We know that:
$\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$

Applying this with $a = m$ and $b = 2$:
$I = \frac{1}{2} \cdot \frac{e^{m\theta}}{m} + \frac{1}{2} \cdot \frac{e^{m\theta}}{m^2 + 4} (m \cos 2\theta + 2 \sin 2\theta) + C$
$I = \frac{e^{m\theta}}{2} \left[ \frac{1}{m} + \frac{m \cos 2\theta + 2 \sin 2\theta}{m^2 + 4} \right] + C$

Now, express $\cos 2\theta$ and $\sin 2\theta$ in terms of $x$:
$\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - x^2}{1 + x^2}$
$\sin 2\theta = \frac{2 \tan\theta}{1 + \tan^2\theta} = \frac{2x}{1 + x^2}$

Also, $\theta = \tan^{-1}x$:
$I = \frac{e^{m \tan^{-1}x}}{2} \left[ \frac{1}{m} + \frac{m(1 - x^2) + 4x}{(m^2 + 4)(1 + x^2)} \right] + C$ (where $C$ is the constant of integration).