\lim_{a\to\infty}\int_{\frac{1}{a}}^{a}\frac{\sqrt{(1+a^{4})x^{2}-(1+x^{4})a^{2}}}{a^{2}(1+x^{2})}-\ln a}

To evaluate the limit:

\[
\lim_{a \to \infty} \int_{\frac{1}{a}}^{a} \frac{\sqrt{(1+a^{4})x^{2} - (1+x^{4})a^{2}}}{a^{2}(1+x^{2})} \, dx - \ln a
\]

let's begin by analyzing the integrand.

### Step 1: Simplifying the integrand
The integrand is:

\[
f(x, a) = \frac{\sqrt{(1+a^{4})x^{2} - (1+x^{4})a^{2}}}{a^{2}(1+x^{2})}
\]

For large \( a \), let's approximate the terms \( (1+a^{4})x^{2} \) and \( (1+x^{4})a^{2} \):

\[
1 + a^4 \approx a^4, \quad \text{and} \quad 1 + x^4 \approx x^4 \quad \text{for large } a \text{ and } x.
\]

Thus, the integrand simplifies to:

\[
f(x, a) \approx \frac{\sqrt{a^4 x^2 - x^4 a^2}}{a^2(1+x^2)} = \frac{a^2\sqrt{x^2 - x^4}}{a^2(1+x^2)} = \frac{\sqrt{x^2 - x^4}}{1+x^2}
\]

Notice that \( x^2 - x^4 = x^2(1 - x^2) \), so the integrand becomes:

\[
f(x, a) \approx \frac{x\sqrt{1 - x^2}}{1+x^2}
\]

### Step 2: Evaluating the integral

Now, we focus on the integral:

\[
\int_{\frac{1}{a}}^{a} \frac{x\sqrt{1 - x^2}}{1+x^2} \, dx
\]

As \( a \to \infty \), the lower limit \( \frac{1}{a} \) approaches 0, and the upper limit \( a \) approaches infinity. However, since the function \( \frac{x\sqrt{1 - x^2}}{1+x^2} \) is nonzero primarily for \( x \) in the interval \( [0, 1] \), the integral over \( [\frac{1}{a}, a] \) reduces effectively to the integral over \( [0, 1] \).

Thus, the integral simplifies to:

\[
\int_{0}^{1} \frac{x\sqrt{1 - x^2}}{1+x^2} \, dx
\]

This is a standard integral, but calculating it exactly involves more complex techniques. Let's approximate it by noting that:

\[
\frac{x\sqrt{1 - x^2}}{1+x^2} \sim x\sqrt{1 - x^2}
\]

Then,

\[
\int_{0}^{1} \frac{x\sqrt{1 - x^2}}{1+x^2} \, dx \approx \int_{0}^{1} x\sqrt{1 - x^2} \, dx
\]

This integral can be computed as:

\[
\int x\sqrt{1 - x^2} \, dx
\]

Use the substitution \( u = 1 - x^2 \), \( du = -2x \, dx \):

\[
-\frac{1}{2} \int \sqrt{u} \, du = -\frac{1}{2} \cdot \frac{2}{3} u^{3/2} = -\frac{1}{3} (1 - x^2)^{3/2} \Big|_0^1 = \frac{1}{3}
\]

Hence,

\[
\int_{0}^{1} x\sqrt{1 - x^2} \, dx = \frac{1}{3}
\]

### Step 3: Considering the logarithmic term

The integral now seems to be a constant as \( a \to \infty \), so the main term to compare with \( \ln a \) is negligible. Thus, the expression \( -\ln a \) dominates the asymptotic behavior.

### Conclusion:

The limit evaluates to:

\[
\lim_{a \to \infty} \left(\frac{1}{3} - \ln a\right) = -\infty
\]

The final result is:

\[
\boxed{-\infty}
\]

Would you like further details or have any questions?

---

**Relative Questions:**

1. How does the approximation of large \( a \) affect the integrand's behavior?
2. Why can we approximate the integrand to focus only on the interval \( [0, 1] \)?
3. What techniques are useful for integrating functions involving square roots and polynomials?
4. How do logarithmic terms behave asymptotically when compared to constants?
5. Can you explain the substitution method used in evaluating the integral?

**Tip:** When solving limits involving integrals, approximations for large or small variables can simplify the problem significantly.

Explore how to evaluate the limit \( \lim_{a \to \infty} \int_{\frac{1}{a}}^{a} \frac{\sqrt{(1+a^{4})x^{2} - (1+x^{4})a^{2}}}{a^{2}(1+x^{2})} \, dx - \ln a \) with detailed asymptotic analysis. This problem delves into the behavior of integrands under large parameter approximations and their impact on the overall limit.

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