## Calculus 8th Edition

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