Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 27

$\sqrt{2x}-ln|1+\sqrt{2x}|+C$

$\int\frac{1}{1+\sqrt{2x}}dx$ let $\sqrt{2x} = (2x)^{\frac{1}{2}} = u$ $du=\frac{1}{2}(2x)^{-\frac{1}{2}}(2)dx$ $du=\frac{1}{\sqrt{2x}}dx$ $du=\frac{1}{u}dx$ $dx=u$ $du$ $\int\frac{1}{1+\sqrt{2x}}dx$ $=\int\frac{u}{1+u}du$ $=\int\frac{1+u-1}{1+u}du$ $=\int(1-\frac{1}{1+u})du$ $=u-ln|1+u|+C$ $=\sqrt{2x}-ln|1+\sqrt{2x}|+C$