Calculus (3rd Edition)

$$\frac{1}{4\sqrt{3}} \ln |\frac{2\sqrt{3}-(x+4)}{2\sqrt{3}+(x+4)}|+C$$
Given $$\int \frac{d x}{x^{2}+8 x+4}$$ Since \begin{aligned} \int \frac{d x}{x^{2}+8 x+4} &=\int \frac{d x}{(x+4)^{2}-12} \end{aligned} Let $$x+4 = \sqrt{12}\sec u\ \ \ \ \ \ dx=\sqrt{12}\sec u\tan udu$$ Then \begin{align*} \int \frac{d x}{x^{2}+8 x+4} &=\int \frac{d x}{(x+4)^{2}-12}\\ &= \int \frac{\sqrt{12}\sec u\tan udu}{12\sec^{2}u-12}\\ &= \int \frac{\sqrt{12}\sec u\tan udu}{12\tan^{2}u}\\ &=\frac{1}{\sqrt{12}} \int \frac{ \sec u du}{ \tan u}\\ &=\frac{1}{\sqrt{12}} \int \frac{ \sec u du}{ \tan u}\\ &=\frac{1}{\sqrt{12}} \int\csc udu\\ &=\frac{1}{\sqrt{12}} \ln | \cot u-\csc u|+C\\ &=\frac{1}{\sqrt{12}}\times-\frac{1}{2} \ln |\frac{\cos u+1}{\cos u - 1}|+C\\ &=\frac{1}{4\sqrt{3}} \ln |\frac{\cos u-1}{\cos u + 1}|+C\\ &=\frac{1}{4\sqrt{3}} \ln |\frac{\frac{\sqrt{12}}{x+4}-1}{\frac{\sqrt{12}}{x+4} + 1}|+C\\ &=\frac{1}{4\sqrt{3}} \ln |\frac{\sqrt{12}-(x+4)}{\sqrt{12}+(x+4)}|+C\\ &=\frac{1}{4\sqrt{3}} \ln |\frac{2\sqrt{3}-(x+4)}{2\sqrt{3}+(x+4)}|+C\\ \end{align*}