The task is to determine whether each given improper integral converges ("сходится") or diverges ("расходится"). I'll analyze each integral step by step to classify them:

Given integrals:

1. $\int_{1}^{+\infty} \frac{\ln x}{\sqrt{x^2 + 1}} \, dx$
2. $\int_{1}^{+\infty} \frac{\ln x}{\sqrt{x}} \, dx$
3. $\int_{1}^{+\infty} \frac{\ln x}{x^2} \, dx$
4. $\int_{1}^{+\infty} \frac{\ln^2 x}{x^2} \, dx$

Analysis:

1. $\int_{1}^{+\infty} \frac{\ln x}{\sqrt{x^2 + 1}} \, dx$

• For large $x$, $\sqrt{x^2 + 1} \approx x$, so the integrand behaves like: $\frac{\ln x}{\sqrt{x^2 + 1}} \sim \frac{\ln x}{x}.$
• The comparison integral is $\int_{1}^{+\infty} \frac{\ln x}{x} \, dx$, which diverges because the logarithmic growth leads to divergence (it is known that $\int_{1}^{+\infty} \frac{1}{x} \, dx$ diverges, and $\ln x$ makes the divergence worse).
• Conclusion: Diverges.

2. $\int_{1}^{+\infty} \frac{\ln x}{\sqrt{x}} \, dx$

• For large $x$, the integrand behaves as: $\frac{\ln x}{\sqrt{x}}.$
• The comparison integral is $\int_{1}^{+\infty} \frac{\ln x}{x^{1/2}} \, dx$. Substituting $u = \sqrt{x}$, this transforms into a divergent form: $\int_{1}^{+\infty} \ln x \cdot x^{-1/2} \, dx \text{ diverges since }\ln x \text{ grows unbounded.}$
• Conclusion: Diverges.

3. $\int_{1}^{+\infty} \frac{\ln x}{x^2} \, dx$

• For large $x$, the integrand behaves like: $\frac{\ln x}{x^2}.$
• The comparison integral is $\int_{1}^{+\infty} \frac{\ln x}{x^2} \, dx$. Integrating by parts ($u = \ln x, dv = x^{-2} dx$): $\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \int -\frac{1}{x^2} dx = -\frac{\ln x}{x} + \frac{1}{x}.$ As $x \to \infty$, this converges because $-\frac{\ln x}{x} \to 0$ and $\frac{1}{x} \to 0$.
• Conclusion: Converges.

4. $\int_{1}^{+\infty} \frac{\ln^2 x}{x^2} \, dx$

• For large $x$, the integrand behaves like: $\frac{\ln^2 x}{x^2}.$
• Similar to the previous case, the comparison integral is $\int_{1}^{+\infty} \frac{\ln^2 x}{x^2} \, dx$. Using the same integration by parts technique ($u = \ln^2 x, dv = x^{-2} dx$): $\int \frac{\ln^2 x}{x^2} \, dx \text{ converges because }\ln^2 x \text{ is still dominated by }x^{-2}.$
• Conclusion: Converges.

Final Classification:

1. $\int_{1}^{+\infty} \frac{\ln x}{\sqrt{x^2 + 1}} \, dx$ → Diverges.
2. $\int_{1}^{+\infty} \frac{\ln x}{\sqrt{x}} \, dx$ → Diverges.
3. $\int_{1}^{+\infty} \frac{\ln x}{x^2} \, dx$ → Converges.
4. $\int_{1}^{+\infty} \frac{\ln^2 x}{x^2} \, dx$ → Converges.

Would you like me to explain any part in more detail? Let me know! 😊

Related Questions:

1. What is an improper integral, and how is it evaluated?
2. How does comparison testing work for improper integrals?
3. Why does logarithmic growth often lead to divergence in integrals?
4. Can you explain how integration by parts works for evaluating $\int \frac{\ln x}{x^2}$?
5. How does behavior analysis at infinity help determine convergence?

Tip:

Always check the asymptotic behavior of the integrand as $x \to \infty$ when determining convergence of improper integrals!