Let's evaluate the given integral:

$I = \int_0^1 \frac{1}{x(2x+5)} \, dx$

First, we need to check if the integral converges, especially near the boundaries of integration.

Step 1: Investigate possible issues

Notice that the integrand contains a factor of $\frac{1}{x}$, and since $x = 0$ is a point of potential concern (it would cause division by zero), we should carefully analyze the behavior near $x = 0$. Let's see if the integral has any improper behavior.

Step 2: Break the integral into simpler parts

We will decompose the integrand using partial fraction decomposition:

$\frac{1}{x(2x+5)} = \frac{A}{x} + \frac{B}{2x+5}$

Multiply both sides by $x(2x + 5)$ to find $A$ and $B$:

$1 = A(2x + 5) + Bx$

Expanding both sides:

$1 = A(2x) + A(5) + Bx$

This gives us:

$1 = (2A + B)x + 5A$

Now, equate the coefficients of like powers of $x$:

• Coefficient of $x$: $2A + B = 0$
• Constant term: $5A = 1$

From $5A = 1$, we find:

$A = \frac{1}{5}$

Substitute $A = \frac{1}{5}$ into $2A + B = 0$:

$2 \times \frac{1}{5} + B = 0 \quad \Rightarrow \quad B = -\frac{2}{5}$

So, the partial fraction decomposition is:

$\frac{1}{x(2x+5)} = \frac{1/5}{x} - \frac{2/5}{2x+5}$

Step 3: Rewrite the integral

Now the integral becomes:

$I = \int_0^1 \left( \frac{1/5}{x} - \frac{2/5}{2x+5} \right) \, dx$

We can split this into two integrals:

$I = \frac{1}{5} \int_0^1 \frac{1}{x} \, dx - \frac{2}{5} \int_0^1 \frac{1}{2x+5} \, dx$

Step 4: Evaluate each integral

1. First integral:

$\int_0^1 \frac{1}{x} \, dx$

This is an improper integral because $\frac{1}{x}$ has a singularity at $x = 0$. To handle this, we approach it as a limit:

$\int_0^1 \frac{1}{x} \, dx = \lim_{\epsilon \to 0^+} \int_\epsilon^1 \frac{1}{x} \, dx$

The antiderivative of $\frac{1}{x}$ is $\ln|x|$, so:

$\int_\epsilon^1 \frac{1}{x} \, dx = \ln(1) - \ln(\epsilon) = -\ln(\epsilon)$

As $\epsilon \to 0^+$, $\ln(\epsilon) \to -\infty$. Therefore, the first integral diverges.

Conclusion

Since the integral $\int_0^1 \frac{1}{x} \, dx$ diverges, the entire integral does not converge. It is divergent due to the singularity at $x = 0$.

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Would you like to explore a different approach, or do you have further questions about this result?

Here are some related questions for further learning:

1. How do improper integrals work, and what is their general form?
2. Can you show how to handle other types of singularities in integrals?
3. What is the difference between absolute and conditional convergence in integrals?
4. How can you determine the convergence of an integral using comparison tests?
5. What are the practical applications of improper integrals in physics?

Tip: When dealing with integrals involving singularities, always check if the integrand has any undefined points or asymptotic behavior, and handle them by approaching as limits.