## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{\ln x}{2}\sqrt{4+(\ln x)^{2}}+2\ln[\ln x+\sqrt{4+(\ln x)^{2}}]+C$
$\displaystyle \int\frac{\sqrt{4+(\ln x)^{2}}}{x}dx=\quad$substitute $\left[\begin{array}{ll} u=\ln x & \\ du=\dfrac{dx}{x} & \end{array}\right]$ $=\displaystyle \int\sqrt{2^{2}+u^{2}}du=$ Table of integrals: $\color{blue}{ 21. \quad \displaystyle \int\sqrt{a^{2}+u^{2}}du=\frac{u}{2}\sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2}\ln(u+\sqrt{a^{2}+u^{2}})+C }$ $=\displaystyle \frac{u}{2}\sqrt{4+u^{2}}+\frac{4}{2}\ln(u+\sqrt{4+u^{2}})+C$ ... bring back x... $=\displaystyle \frac{\ln x}{2}\sqrt{4+(\ln x)^{2}}+2\ln[\ln x+\sqrt{4+(\ln x)^{2}}]+C$