Answer
$$\displaystyle{\int\frac{x}{\sqrt{x^2-7}}dx=\sqrt{x^2-7}+C}\\$$
Work Step by Step
$\displaystyle{I=\int\frac{x}{\sqrt{x^2-7}}dx}\\$
$\displaystyle \left[\begin{array}{ll} t=\sqrt7\sec\theta & t^2=7\sec^2\theta \\ & \\ \frac{dt}{d\theta}=\sqrt7\sec\theta\tan\theta & dt=\sqrt7\sec\theta\tan\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int\frac{\sqrt7\sec\theta}{\sqrt{7\sec^2\theta-7}}\sqrt7\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int\frac{\sqrt7\sec\theta}{\sqrt{7\left(\sec^2\theta-1\right)}}\sqrt7\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int{\sqrt7\sec^2\theta}\ d\theta}\\ \displaystyle{I=\sqrt7\int\sec^2\theta\ d\theta}\\ \displaystyle{I=\sqrt7\tan\theta+C}\\$
$\tan\theta=\frac{\sqrt{x^2-7}}{ \sqrt7}\\$
$\displaystyle{I=\sqrt7\times\frac{\sqrt{x^2-7}}{ \sqrt7}+C}\\ \displaystyle{I=\sqrt{x^2-7}+C}\\$
