1. **State the problem:** Evaluate the limit $$\lim_{x \to \infty} \left( \frac{x^3}{3} \cdot 40 \cdot \frac{\pi}{4} - \arctan(x) + \frac{1}{3} \left( \frac{x^2}{2} + 0.5 \log(1 + x^2) \right) \right)$$ 2. **Rewrite the expression for clarity:** $$\lim_{x \to \infty} \left( \frac{40 \pi}{12} x^3 - \arctan(x) + \frac{1}{3} \left( \frac{x^2}{2} + \frac{1}{2} \log(1 + x^2) \right) \right)$$ 3. **Simplify constants:** $$\frac{40 \pi}{12} = \frac{10 \pi}{3}$$ So the expression becomes: $$\lim_{x \to \infty} \left( \frac{10 \pi}{3} x^3 - \arctan(x) + \frac{1}{3} \left( \frac{x^2}{2} + \frac{1}{2} \log(1 + x^2) \right) \right)$$ 4. **Analyze each term as $x \to \infty$:** - $\frac{10 \pi}{3} x^3$ grows without bound to $+\infty$. - $\arctan(x) \to \frac{\pi}{2}$ (finite constant). - $\frac{1}{3} \left( \frac{x^2}{2} + \frac{1}{2} \log(1 + x^2) \right)$ grows without bound, dominated by $\frac{x^2}{6}$ term. 5. **Dominant term:** The cubic term $\frac{10 \pi}{3} x^3$ dominates the growth. 6. **Conclusion:** Since the leading term tends to $+\infty$, the whole expression tends to $+\infty$. **Final answer:** $$\lim_{x \to \infty} \left( \frac{x^3}{3} \cdot 40 \cdot \frac{\pi}{4} - \arctan(x) + \frac{1}{3} \left( \frac{x^2}{2} + 0.5 \log(1 + x^2) \right) \right) = +\infty$$