We need to evaluate the integral:

$I = \int_{0}^{1} \frac{12}{4x^2 + 5x + 1} \, dx$

Step 1: Complete the square for the quadratic in the denominator

The expression in the denominator is $4x^2 + 5x + 1$. To simplify, let's complete the square.

First, write $4x^2 + 5x + 1$ in the form $a(x - h)^2 + k$.

1. Factor out 4 from the $x^2$ and $x$ terms:

   $4x^2 + 5x + 1 = 4\left(x^2 + \frac{5}{4}x \right) + 1$

2. To complete the square, take half of the coefficient of $x$ inside the parentheses, square it, and add/subtract the value. Half of $\frac{5}{4}$ is $\frac{5}{8}$, and squaring it gives $\frac{25}{64}$. So, we add and subtract $\frac{25}{64}$ inside the parentheses:

   $4\left(x^2 + \frac{5}{4}x + \frac{25}{64} - \frac{25}{64} \right) + 1$

3. This simplifies to:

   $4\left(\left(x + \frac{5}{8}\right)^2 - \frac{25}{64}\right) + 1$

4. Distribute the 4 and simplify:

   $4\left(x + \frac{5}{8}\right)^2 - \frac{100}{64} + 1 = 4\left(x + \frac{5}{8}\right)^2 - \frac{100}{64} + \frac{64}{64} = 4\left(x + \frac{5}{8}\right)^2 - \frac{36}{64}$

   $= 4\left(x + \frac{5}{8}\right)^2 - \frac{9}{16}$

Thus, the denominator becomes:

$4x^2 + 5x + 1 = 4\left(x + \frac{5}{8}\right)^2 + \frac{7}{16}$

Step 2: Use substitution

Let $u = x + \frac{5}{8}$. Then $du = dx$, and we can transform the limits accordingly:

• When $x = 0$, $u = \frac{5}{8}$,
• When $x = 1$, $u = \frac{13}{8}$.

The integral becomes:

$I = \int_{\frac{5}{8}}^{\frac{13}{8}} \frac{12}{4u^2 + \frac{7}{16}} \, du$

Factor out the constants:

$I = \int_{\frac{5}{8}}^{\frac{13}{8}} \frac{12}{4\left(u^2 + \frac{7}{64}\right)} \, du$

$I = 3 \int_{\frac{5}{8}}^{\frac{13}{8}} \frac{1}{u^2 + \left(\frac{\sqrt{7}}{8}\right)^2} \, du$

Step 3: Apply the standard integral formula

This is a standard integral of the form:

$\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$

Here, $a = \frac{\sqrt{7}}{8}$. So the integral becomes:

$I = \frac{3 \times 8}{\sqrt{7}} \left[\arctan\left(\frac{8u}{\sqrt{7}}\right)\right]_{\frac{5}{8}}^{\frac{13}{8}}$

Step 4: Evaluate the limits

Substitute the limits $u = \frac{13}{8}$ and $u = \frac{5}{8}$:

$I = \frac{24}{\sqrt{7}} \left[\arctan\left(\frac{8 \times \frac{13}{8}}{\sqrt{7}}\right) - \arctan\left(\frac{8 \times \frac{5}{8}}{\sqrt{7}}\right)\right]$

$I = \frac{24}{\sqrt{7}} \left[\arctan\left(\frac{13}{\sqrt{7}}\right) - \arctan\left(\frac{5}{\sqrt{7}}\right)\right]$

This is the exact value of the integral.

Would you like me to compute the approximate numerical value for you? Let me know if you want further details or have questions!

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Related questions:

1. What is the arctangent function and how does it apply in integration?
2. Can you explain how to complete the square for different types of quadratic equations?
3. How do you evaluate definite integrals involving trigonometric functions?
4. What are standard integral formulas, and how can they simplify problems?
5. How does substitution simplify complex integrals?

Tip: Completing the square is a powerful technique to simplify integrals involving quadratic expressions.