Solutions - 0050-04

(1) Complete the square:

$$\begin{array}{c}{x}^2+4x+13\gg \gg x^2+4x+{\left(\frac{4}{2}\right)}^2-{\left(\frac{4}{2}\right)}^2+13\\ \\ \gg \gg {(x+2)}^2+9\end{array}$$

---

(2) Substitute $x+2=3\tan \theta$ and thus $dx=3{\sec }^2\theta d\theta$:

$$\begin{array}{c}\gg \gg \int \frac{dx}{\sqrt{{(x+2)}^2+9}}\gg \gg \int \frac{3{\sec }^2\theta d\theta }{\sqrt{9{\tan }^2\theta +9}}\\ \\ \gg \gg \int \sec \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |+C\end{array}$$

---

(3) Convert back to terms of $x$:

First draw a triangle expressing $\tan \theta ={\displaystyle \frac{x+2}{3}}$:

It follows that $\sec \theta ={\displaystyle \frac{\sqrt{{(x+2)}^2+9}}{3}}$. Then:

$$\begin{array}{ccc}& \ln |\tan \theta +\sec \theta |+C& \\ & & \\ & \gg \gg \ln |\frac{x+2}{3}+\frac{\sqrt{{(x+2)}^2+9}}{3}|+C& \\ & & \\ & \gg \gg \ln |x+2+\sqrt{{(x+2)}^2+9}|+C& \text{(Note A)}\end{array}$$

---

Note A: Using log rules, the denominator $3$ can be brought out as $-\ln 3$ which can be “absorbed” into the constant $C$.