## Calculus 8th Edition

$$\displaystyle{\int\frac{1}{\sqrt{x^2+2x+5}}dx=\ln\left|\frac{\sqrt{ x^2+2x+5}+x+1}{2}\right|+C}$$
$\displaystyle{I=\int\frac{1}{\sqrt{x^2+2x+5}}dx}\\ \displaystyle{I=\int\frac{1}{\sqrt{\left(x+1\right)^2+4}}dx}\\$ $\displaystyle \left[\begin{array}{ll} x+1=2\tan\theta & \left(x+1\right)^2=4\tan^2\theta \\ & \\ \frac{dx}{d\theta}=2\sec^2\theta & dx=2\sec^2\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int\frac{1}{\sqrt{4\tan^2\theta+4}}2\sec^2\theta\ d\theta}\\ \displaystyle{I=\int\frac{1}{\sqrt{4\left(\tan^2\theta+1\right)}}2\sec^2\theta\ d\theta}\\ \displaystyle{I=\int\frac{1}{2\sec\theta}2\sec^2\theta\ d\theta}\\ \displaystyle{I=\int\sec\theta\ d\theta}\\ \displaystyle{I=\ln\left|\sec\theta+\tan\theta\right|+C}\\$ $\sec\theta=\frac{\sqrt{ x^2+2x+5}}{2}\\ \tan\theta=\frac{x+1}{2}\\$ $\displaystyle{I=\ln\left|\frac{\sqrt{ x^2+2x+5}}{2}+\frac{x+1}{2}\right|+C}\\ \displaystyle{I=\ln\left|\frac{\sqrt{ x^2+2x+5}+x+1}{2}\right|+C}\\$