Evaluate the integral. \[ \int_{0}^{1} \frac{12}{4 x^{2}+5 x+1} d x \]

To evaluate the integral \[ I = \int_{0}^{1} \frac{12}{4 x^{2}+5 x+1} \, dx, \] we start by factoring the quadratic expression in the denominator. The quadratic \(4x^2 + 5x + 1\) can be factored using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} = \frac{-5 \pm \sqrt{25 - 16}}{8} = \frac{-5 \pm 3}{8}. \] This gives us the roots: \[ x_1 = \frac{-2}{8} = -\frac{1}{4}, \quad x_2 = \frac{-8}{8} = -1. \] Thus, we can write \[ 4x^2 + 5x + 1 = 4\left(x + \frac{1}{4}\right)\left(x + 1\right). \] We can proceed with partial fraction decomposition: \[ \frac{12}{4(x + \frac{1}{4})(x + 1)} = \frac{A}{x + \frac{1}{4}} + \frac{B}{x + 1}. \] Multiplying through by the denominator gives: \[ 12 = A\left(x + 1\right) + B\left(x + \frac{1}{4}\right). \] To find coefficients \(A\) and \(B\), we can choose convenient values of \(x\). 1. Let \(x = -1\): \[ 12 = A(0) + B\left(-1 + \frac{1}{4}\right) \quad \Rightarrow \quad 12 = B\left(-\frac{3}{4}\right) \quad \Rightarrow \quad B = -16. \] 2. Let \(x = -\frac{1}{4}\): \[ 12 = A\left(-\frac{1}{4} + 1\right) + B(0) \quad \Rightarrow \quad 12 = A\left(\frac{3}{4}\right) \quad \Rightarrow \quad A = 16. \] Thus, the partial fraction decomposition is: \[ \frac{12}{4x^2 + 5x + 1} = \frac{16}{x + \frac{1}{4}} - \frac{16}{x + 1}. \] Now we can rewrite the integral: \[ I = \int_{0}^{1} \left(\frac{16}{x + \frac{1}{4}} - \frac{16}{x + 1}\right) \, dx. \] This separates into two integrals: \[ I = 16 \int_{0}^{1} \frac{1}{x + \frac{1}{4}} \, dx - 16 \int_{0}^{1} \frac{1}{x + 1} \, dx. \] Calculating the first integral: \[ \int \frac{1}{x + \frac{1}{4}} \, dx = \ln \left|x + \frac{1}{4}\right|. \] Evaluating from 0 to 1: \[ \left[ \ln \left|x + \frac{1}{4}\right| \right]_{0}^{1} = \ln \left(1 + \frac{1}{4}\right) - \ln \left(\frac{1}{4}\right) = \ln \left(\frac{5}{4}\right) + \ln 4 = \ln 5. \] For the second integral: \[ \int \frac{1}{x + 1} \, dx = \ln |x + 1|. \] Evaluating from 0 to 1: \[ \left[ \ln |x + 1| \right]_{0}^{1} = \ln 2 - \ln 1 = \ln 2. \] Putting it all together: \[ I = 16 \left(\ln 5 - \ln 2\right) = 16 \ln \left(\frac{5}{2}\right). \] Thus, the final value of the integral is: \[ \boxed{16 \ln \left(\frac{5}{2}\right)}. \]