Let F be the antiderivative of 1/(x² + 4x + 13) with F(-2) = 0…

Step 1. Complete the square. $x^2 + 4x + 13 = (x + 2)^2 + 9$.

Step 2. Use the standard form. With $u = x + 2$ and $a = 3$, the integral is $\dfrac{1}{3}\tan^{-1}\dfrac{x + 2}{3} + C$. So $b = \dfrac{1}{3} = 0.3333$ and $s = 3$.

Step 3. The constant. At $x = -2$, $\tan^{-1}0 = 0$, so $C = 0$ already.

Step 4. Evaluate. $F(1) = \dfrac{1}{3}\tan^{-1}\dfrac{3}{3} = \dfrac{1}{3} \times \dfrac{\pi}{4} = 0.2618$.