Calculus (3rd Edition)

$\frac{1}{6}(4x-1)^{\frac{3}{2}}+C$
Substituting u=4x-1 so that $dx=\frac{du}{4}$ We get $\int \sqrt {4x-1}dx=\int u^{\frac{1}{2}}\frac{du}{4}$ $=\frac{1}{4}\int u^{\frac{1}{2}}du=\frac{1}{4}\times\frac{2}{3}u^{\frac{3}{2}}+C$ $=\frac{1}{6}u^{\frac{3}{2}}+C$ $=\frac{1}{6}(4x-1)^{\frac{3}{2}}+C$