The given integral is:

$\int_{0}^{1} \frac{dx}{x^2}.$

Step-by-step solution:

1. Rewrite $\frac{1}{x^2}$ as $x^{-2}$. $\int_{0}^{1} \frac{dx}{x^2} = \int_{0}^{1} x^{-2} \, dx.$

2. Use the power rule for integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1}, \quad \text{for } n \neq -1.$

   Here $n = -2$, so: $\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -\frac{1}{x}.$

3. Apply the limits of integration from $0$ to $1$: $\int_{0}^{1} x^{-2} \, dx = \left[-\frac{1}{x}\right]_{0}^{1}.$

4. Evaluate the expression at the bounds:

   • At $x = 1$: $-\frac{1}{1} = -1.$
   • At $x = 0$, the term $-\frac{1}{x}$ becomes undefined because division by zero is not allowed.

Conclusion:

The integral diverges as $x \to 0$, meaning it is not finite. Therefore:

$\int_{0}^{1} \frac{dx}{x^2} \quad \text{is divergent.}$

Do you want further clarification or related examples?

Related Questions:

1. What is the difference between convergent and divergent integrals?
2. How do you evaluate improper integrals with different types of singularities?
3. What happens if we change the limits of this integral to $[a, 1]$ with $a > 0$?
4. Can you explain the significance of a divergent integral in physics or engineering?
5. How do you handle integrals that have infinite bounds?

Tip:

For integrals that involve undefined points (like division by zero or $\infty$), always check for convergence using limits!