## Calculus: Early Transcendentals 8th Edition

$\displaystyle \ln|\frac{\sqrt{x^{2}+1}-1}{x}|+C$
There is a $\sqrt{x^{2}+a^{2}}$ term in the integrand, so we try the trigonometric substitution $\displaystyle \int\frac{dx}{x\sqrt{x^{2}+1}}=\left[\begin{array}{ll} x=\tan t & \\ dx=\sec^{2}tdt & \end{array}\right],\displaystyle \qquad \left[\begin{array}{l} \sec^{2}t=\tan^{2}t+1\\ \sec t=\sqrt{x^{2}+1} \end{array}\right]$ $=\displaystyle \int\frac{\sec^{2}tdt}{\tan t\sec t}\qquad$... simplify (cancel one sec$...)$ $=\displaystyle \int\frac{\sec t}{\tan t}dt$ $=\displaystyle \int\csc tdt$ $=\ln|\csc t-\cot t|+C$ ... bring back x $\left[\begin{array}{ll} \csc t=\dfrac{\sec t}{\tan t}=\dfrac{\sqrt{x^{2}+1}}{x}, & \cot t=\dfrac{1}{x}\\ & \end{array}\right]$ $=\displaystyle \ln|\frac{\sqrt{x^{2}+1}}{x}-\frac{1}{x}|+C$ $=\displaystyle \ln|\frac{\sqrt{x^{2}+1}-1}{x}|+C$