Evaluate : \(\int\frac{1}{4x^2-20x+17}\,dx\)

Answer: \(\frac{1}{8\sqrt{2}}\log\left|\frac{2x-5-2\sqrt{2}}{2x-5+2\sqrt{2}}\right|+c\)

\(4x^2-20x+17=4\left(x^2-5x+\frac{17}{4}\right)=4\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}+\frac{17}{4}\right]=4\left[\left(x-\frac{5}{2}\right)^2-(\sqrt{2})^2\right]\)

\(I=\frac{1}{4}\int\frac{dx}{\left(x-\frac{5}{2}\right)^2-(\sqrt{2})^2}\)

Using \(\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+c\):

\(I=\frac{1}{4}\cdot\frac{1}{2\sqrt{2}}\log\left|\frac{x-\frac{5}{2}-\sqrt{2}}{x-\frac{5}{2}+\sqrt{2}}\right|+c=\frac{1}{8\sqrt{2}}\log\left|\frac{2x-5-2\sqrt{2}}{2x-5+2\sqrt{2}}\right|+c\)