The following integral converges. Evaluate the integral without using tables. \( \int_{-1}^{\infty} \frac{4}{x^{2}+5 x+6} d x \) \( \int_{-1}^{\infty} \frac{4}{x^{2}+5 x+6} d x=\square \) (Type an exact answer.)

To evaluate the integral \( \int_{-1}^{\infty} \frac{4}{x^{2}+5 x+6} d x \), we start by factoring the quadratic in the denominator. The expression \( x^2 + 5x + 6 \) factors as \( (x+2)(x+3) \). So we rewrite the integral: \[ \int_{-1}^{\infty} \frac{4}{(x+2)(x+3)} \, dx \] Next, we can use partial fraction decomposition: \[ \frac{4}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3} \] Multiplying through by \( (x+2)(x+3) \) gives: \[ 4 = A(x+3) + B(x+2) \] Setting \( x = -2 \): \[ 4 = A(1) + B(0) \implies A = 4 \] Setting \( x = -3 \): \[ 4 = A(0) + B(-1) \implies B = -4 \] Now we have: \[ \frac{4}{(x+2)(x+3)} = \frac{4}{x+2} - \frac{4}{x+3} \] Thus, the integral becomes: \[ \int_{-1}^{\infty} \left( \frac{4}{x+2} - \frac{4}{x+3} \right) dx \] We can evaluate these integrals separately. 1. \( \int \frac{4}{x+2} \, dx = 4 \ln |x+2| \) 2. \( \int \frac{4}{x+3} \, dx = 4 \ln |x+3| \) So we have: \[ \int_{-1}^{\infty} \frac{4}{x+2} \, dx - \int_{-1}^{\infty} \frac{4}{x+3} \, dx \] Evaluating the limits, first for \( \int_{-1}^\infty \frac{4}{x+2} \, dx \): \[ \lim_{b \to \infty} \left[ 4 \ln |b+2| - 4 \ln |1| \right] = \lim_{b \to \infty} 4 \ln(b+2) \to \infty \] And for \( \int_{-1}^\infty \frac{4}{x+3} \, dx \): \[ \lim_{b \to \infty} \left[ 4 \ln |b+3| - 4 \ln |0| \right] \to \infty \] Both integrals diverge as \( b \to \infty \). However, when taking the limit for the definite bounds from \( -1 \) to \( a \) where \( a \) approaches infinity: \[ \int_{-1}^{a} \left( \frac{4}{x+2} - \frac{4}{x+3} \right) dx \] Evaluating from \(-1\) to \(a\), simplifies to \(4 (\ln(a+2) - \ln(1)) - 4 (\ln(a+3) - \ln(0))\) simplifies to \(4 \ln\left(\frac{a+2}{a+3}\right)\), recognizing that this approaches \(4\ln\left(1\right) = 0\) as \(a\) approaches infinity. Thus, the definite integral converges to: \[ \int_{-1}^{\infty} \frac{4}{(x+2)(x+3)} \, dx = 2 \ln(2) \] So the final answer is: \[ \int_{-1}^{\infty} \frac{4}{x^{2}+5 x+6} d x = 2 \ln(2) \] \(\int_{-1}^{\infty} \frac{4}{x^{2}+5 x+6} d x = 2 \ln(2)\)