Answer
$$\ln \left|\frac{x+2}{3}+\frac{\sqrt{x^{2}+4 x+13}}{3}\right|+C $$
or
$$ \ln \left|x+2+\sqrt{x^{2}+4 x+13}\right|+C $$
Work Step by Step
Given $$\int \frac{d x}{\sqrt{x^{2}+4 x+13}}$$ Since \begin{align*} x^{2}+4 x+13&= (x+2)^2-4+13\\ &=(x+2)^2-9 \end{align*} Let $$x+2=3\sec u\ \ \ \to \ \ \ \ dx=3\sec u\tan udu$$ Then \begin{align*} \int \frac{d x}{\sqrt{x^{2}+4 x+13}}&=\int \frac{d x}{\sqrt{(x+2)^2-9}}\\ &= \int \frac{3\sec u\tan udu}{\sqrt{9\sec^2 u-9}}\\ &= \int \frac{3\sec u\tan udu}{\sqrt{9\tan ^2u}}\\ &= \int \sec u du\\ &=\ln |\sec u+\tan u|+C\\ &= \ln \left|\frac{x+2}{3}+\frac{\sqrt{x^{2}+4 x+13}}{3}\right|+C
\end{align*}
or, we can write:
$$ \ln \left|x+2+\sqrt{x^{2}+4 x+13}\right|+C $$
