[Cambridge/9233/2004] 写出常数 $A$ 和 $B$，使得对于所有 $x$ 的值，
$6x + 16 = A(2x + 4) + B.$

1. Find the constants $A$ and $B$ from the earlier part to rewrite the numerator.

By comparing coefficients in $6x + 16 = A(2x + 4) + B$:
$2A = 6 \implies A = 3$
$4A + B = 16 \implies 3(4) + B = 16 \implies B = 4$

2. Express the integral as the sum of two fractions by substituting $A$ and $B$, and complete the square for the denominator.

$\int_{-2}^{1} \frac{6x + 16}{x^2 + 4x + 13} \, dx = \int_{-2}^{1} \frac{3(2x + 4) + 4}{x^2 + 4x + 13} \, dx$
$= \int_{-2}^{1} \left( \frac{3(2x + 4)}{x^2 + 4x + 13} + \frac{4}{(x+2)^2 + 9} \right) \, dx$

3. Integrate the expression using standard formulas for $\ln$ and $\tan^{-1}$.

$= \left[ 3 \ln (x^2 + 4x + 13) + \frac{4}{3} \tan^{-1} \left( \frac{x+2}{3} \right) \right]_{-2}^{1}$

4. Substitute the upper and lower limits into the integrated expression.

$= \left( 3 \ln (1^2 + 4(1) + 13) + \frac{4}{3} \tan^{-1} \left( \frac{1+2}{3} \right) \right) - \left( 3 \ln ((-2)^2 + 4(-2) + 13) + \frac{4}{3} \tan^{-1} \left( \frac{-2+2}{3} \right) \right)$
$= \left( 3 \ln 18 + \frac{4}{3} \tan^{-1}(1) \right) - \left( 3 \ln 9 + \frac{4}{3} \tan^{-1}(0) \right)$
$= 3 \ln 18 + \frac{4}{3} \left(\frac{\pi}{4}\right) - 3 \ln 9 - 0$

5. Simplify the resulting expression using logarithm properties to find the exact value.

$= 3 \ln 18 - 3 \ln 9 + \frac{\pi}{3}$
$= 3 \ln \left( \frac{18}{9} \right) + \frac{\pi}{3}$
$= 3 \ln 2 + \frac{\pi}{3}$