We are tasked with evaluating the following definite integral:

$I = \int_{0}^{2} \frac{2}{x^2 + 2x + 5} \, dx$

Step 1: Completing the square

To simplify the expression in the denominator, we complete the square for $x^2 + 2x + 5$. We focus on the quadratic part $x^2 + 2x$ and rewrite it as:

$x^2 + 2x = (x+1)^2 - 1$

Thus, the denominator becomes:

$x^2 + 2x + 5 = (x+1)^2 - 1 + 5 = (x+1)^2 + 4$

Step 2: Substituting $u = x + 1$

Let’s make the substitution $u = x + 1$, so $du = dx$. When $x = 0$, we have $u = 1$, and when $x = 2$, we have $u = 3$. Therefore, the integral becomes:

$I = \int_{1}^{3} \frac{2}{u^2 + 4} \, du$

Step 3: Using a standard integral formula

The integral $\int \frac{du}{u^2 + a^2}$ is a standard integral, with the result:

$\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right)$

In our case, $a = 2$. Thus, the integral becomes:

$I = \int_{1}^{3} \frac{2}{u^2 + 4} \, du = 2 \times \frac{1}{2} \left[ \tan^{-1} \left( \frac{u}{2} \right) \right]_{1}^{3}$

Simplifying:

$I = \left[ \tan^{-1} \left( \frac{u}{2} \right) \right]_{1}^{3}$

Step 4: Evaluating the limits

Now we evaluate the expression at the upper and lower limits:

$I = \tan^{-1} \left( \frac{3}{2} \right) - \tan^{-1} \left( \frac{1}{2} \right)$

Step 5: Final answer

Thus, the value of the integral is:

$I = \tan^{-1} \left( \frac{3}{2} \right) - \tan^{-1} \left( \frac{1}{2} \right)$

This is the exact value of the integral. You can approximate the result using a calculator:

$I \approx 0.9828 - 0.4636 = 0.5192$

So, the approximate value of the integral is $0.5192$.

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Would you like more details on any of the steps, or do you have any questions? Here are some related questions for further exploration:

1. What is the general method for solving integrals involving quadratic expressions in the denominator?
2. How do we handle integrals with substitutions to simplify the integrand?
3. What are other techniques for evaluating definite integrals involving trigonometric substitutions?
4. How can we use a calculator to compute inverse tangent values more accurately?
5. Can we apply similar methods to integrals of higher-order polynomials in the denominator?

Tip: When faced with quadratic expressions in the denominator, completing the square often simplifies the integrand and makes it possible to use standard inverse trigonometric identities.