Answer
$$\displaystyle{\int \sqrt{x^2+2x}\ dx=\frac{1}{2}\left(x+1\right)\left(\sqrt{x^2+2x}\right)-\frac{1}{2}\ln\left|\left(x+1\right)+\sqrt{x^2+2x}\right|+C}$$
Work Step by Step
$\displaystyle{I=\int \sqrt{x^2+2x}\ dx}\\ \displaystyle{I=\int \sqrt{\left(x+1\right)^2-1}\ dx}\\$
$\displaystyle \left[\begin{array}{ll} x+1=\sec\theta & \left(x+1\right)^2=\sec^2\theta \\ & \\ \frac{dx}{d\theta}=\sec\theta\tan\theta & dx=\sec\theta\tan\theta d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int \sqrt{\sec^2\theta-1}\sec\theta\tan\theta\ d\theta}\\\displaystyle{I=\int \tan\theta\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int\sec\theta\tan^2\theta\ d\theta}\\ \displaystyle{I=\int\sec\theta\left(\sec^2\theta-1\right)\ d\theta}\\ \displaystyle{I=\int\sec^3\theta-\sec\theta\ d\theta}\\ \displaystyle{I=\int\sec^3\theta\ d\theta-\int \sec\theta\ d\theta}\\$ Integrating $\sec^3\theta$
$\displaystyle \left[\begin{array}{ll} u=\sec\theta & dv=\sec^2\theta \\ & \\ du=\sec\theta\tan\theta & v=\tan\theta \end{array}\right]$ Integration by parts
$\displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta\ d\theta}\\ \displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec\theta\left(\sec^2\theta-1\right)\ d\theta}\\ \displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec^3\theta-\sec\theta\ d\theta}\\ \displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec^3\theta\ d\theta+\int \sec\theta\ d\theta}\\ \displaystyle{2\int\sec^3\theta\ d\theta=\sec\theta\tan\theta+\int \sec\theta\ d\theta}\\ \displaystyle{2\int\sec^3\theta\ d\theta=\sec\theta\tan\theta+\ln\left|\sec \theta+\tan \theta\right|+C}\\ \displaystyle{\int\sec^3\theta\ d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec \theta+\tan \theta\right|+C}\\$
$ \displaystyle{I=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec \theta+\tan \theta\right|-\int \sec\theta\ d\theta}\\ \displaystyle{I=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec \theta+\tan \theta\right|-\ln\left|\sec \theta+\tan \theta\right|+C}\\ \displaystyle{I=\frac{1}{2}\sec\theta\tan\theta-\frac{1}{2}\ln\left|\sec \theta+\tan \theta\right|+C}\\ $
$\sec\theta=x+1\\ \tan\theta=\sqrt{x^2+2x}\\ $
$\displaystyle{I=\frac{1}{2}\left(x+1\right)\left(\sqrt{x^2+2x}\right)-\frac{1}{2}\ln\left|\left(x+1\right)+\sqrt{x^2+2x}\right|+C}\\ $
