[m] ∫ \frac{1}{\sqrt{4x^2-x+4}}dx=\frac{1}{2} ∫ \frac{1}{\sqrt{(x-\frac{1}{8})^2+\frac{63}{64}}}dx=\frac{1}{2} ∫ \frac{d(x-\frac{1}{8})}{\sqrt{(x-\frac{1}{8})^2+\frac{63}{64}}}=\frac{1}{2}ln|(x-\frac{1}{8})+\sqrt{(x^2-\frac{1}{4}+1)}|+C [/m]
По формуле: