To evaluate the integral

$\int \frac{\sqrt{x^2 + 16}}{x^4} \, dx$

we will proceed step by step.

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Step 1: Simplify the integrand

The given integrand is:

$\frac{\sqrt{x^2 + 16}}{x^4}.$

Rewrite it as:

$\frac{\sqrt{x^2 + 16}}{x^4} = x^{-4} \cdot \sqrt{x^2 + 16}.$

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Step 2: Substitution for simplification

Since $\sqrt{x^2 + 16}$ involves $x^2 + 16$, a good substitution is:

$x = 4 \tan \theta \quad \text{so that} \quad x^2 + 16 = 16 \sec^2 \theta.$

Differentiating $x = 4 \tan \theta$, we get:

$dx = 4 \sec^2 \theta \, d\theta.$

Substituting $x$ into $x^{-4}$, we get:

$x^2 = 16 \tan^2 \theta, \quad x^{-4} = \frac{1}{(16 \tan^2 \theta)^2} = \frac{1}{256 \tan^4 \theta}.$

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Step 3: Substitute into the integral

Replace the terms in the integral using the substitution:

$\sqrt{x^2 + 16} = \sqrt{16 \sec^2 \theta} = 4 \sec \theta,$ $dx = 4 \sec^2 \theta \, d\theta.$

Thus, the integral becomes:

$\int \frac{\sqrt{x^2 + 16}}{x^4} \, dx = \int \frac{4 \sec \theta}{256 \tan^4 \theta} \cdot 4 \sec^2 \theta \, d\theta.$

Simplify:

$\int \frac{\sqrt{x^2 + 16}}{x^4} \, dx = \frac{16}{256} \int \frac{\sec^3 \theta}{\tan^4 \theta} \, d\theta = \frac{1}{16} \int \frac{\sec^3 \theta}{\tan^4 \theta} \, d\theta.$

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Step 4: Rewrite in terms of trigonometric identities

We know that:

\sec^2 \theta = 1 + \tan^2 \theta \quad \text{and} \quad \tan^2 \theta = u \quad \text{(substitute u = \tan \theta)}.

Let’s transform $\int \frac{\sec^3 \theta}{\tan^4 \theta}$ into a solvable form and evaluate the integral.

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