Let $a$ be a real number, $a>1.$ Find the real numbers $b \ge 1$ such that $$\lim_{x \to \infty} \int\limits_0^x (1+t^a)^{-b} \mathrm{d}t=1.$$
1. We start with the given integral and its limit:
   \[
   \lim_{x \to \infty} \int_0^x (1+t^a)^{-b} \, \mathrm{d}t = 1.
   \]

2. To simplify the integral, we use the substitution \( t^a = \frac{1}{s} - 1 \). This implies \( t = \left( \frac{1}{s} - 1 \right)^{1/a} \) and \( \mathrm{d}t = -\frac{1}{a} \left( \frac{1}{s} - 1 \right)^{\frac{1}{a} - 1} \frac{1}{s^2} \, \mathrm{d}s \).

3. Changing the limits of integration accordingly, when \( t \) goes from \( 0 \) to \( \infty \), \( s \) goes from \( 1 \) to \( 0 \). Thus, the integral becomes:
   \[
   \int_0^\infty (1+t^a)^{-b} \, \mathrm{d}t = \int_1^0 \left( 1 + \left( \frac{1}{s} - 1 \right) \right)^{-b} \left( -\frac{1}{a} \left( \frac{1}{s} - 1 \right)^{\frac{1}{a} - 1} \frac{1}{s^2} \right) \, \mathrm{d}s.
   \]

4. Simplifying the integrand:
   \[
   \int_1^0 \left( \frac{1}{s} \right)^{-b} \left( -\frac{1}{a} \left( \frac{1}{s} - 1 \right)^{\frac{1}{a} - 1} \frac{1}{s^2} \right) \, \mathrm{d}s = \frac{1}{a} \int_0^1 s^{b - \frac{1}{a} - 1} (1 - s)^{\frac{1}{a} - 1} \, \mathrm{d}s.
   \]

5. Recognizing the integral as a Beta function \( B(x, y) \):
   \[
   \frac{1}{a} \int_0^1 s^{b - \frac{1}{a} - 1} (1 - s)^{\frac{1}{a} - 1} \, \mathrm{d}s = \frac{1}{a} B\left( b - \frac{1}{a}, \frac{1}{a} \right).
   \]

6. The Beta function \( B(x, y) \) is defined as:
   \[
   B(x, y) = \int_0^1 t^{x-1} (1-t)^{y-1} \, \mathrm{d}t = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.
   \]

7. Therefore, we have:
   \[
   \frac{1}{a} B\left( b - \frac{1}{a}, \frac{1}{a} \right) = \frac{1}{a} \frac{\Gamma\left( b - \frac{1}{a} \right) \Gamma\left( \frac{1}{a} \right)}{\Gamma\left( b \right)}.
   \]

8. Given that the integral equals 1, we set:
   \[
   \frac{1}{a} \frac{\Gamma\left( b - \frac{1}{a} \right) \Gamma\left( \frac{1}{a} \right)}{\Gamma\left( b \right)} = 1.
   \]

9. Simplifying, we get:
   \[
   \frac{\Gamma\left( b - \frac{1}{a} \right) \Gamma\left( \frac{1}{a} \right)}{a \Gamma\left( b \right)} = 1.
   \]

10. Using the property of the Gamma function \( \Gamma(x+1) = x \Gamma(x) \), we find:
    \[
    b - \frac{1}{a} = 1 \implies b = \frac{1}{a} + 1.
    \]

Conclusion:
\[
\boxed{b = \frac{1}{a} + 1}
\]