$\int\limits_{2}^{4} \frac{3x-1}{x^{2}-4x+8} dx$=$\frac{3}{2}\int\limits_{2}^{4}\frac{d(x^2-4x+8)}{x^2-4x+8}(I)+5\int\limits_{2}^{4}\frac{dx}{x^2-4x+8}(I_{2})$
$=\frac{3}{2}\ln\left| {x^2-4x+8} \right| [2-4]+I_{2}$
$I_{2}$ $=5\int\limits_{2}^{4}\frac{dx}{(x-2)^2+4}$ Đặt $x-2=2tant$ $dx=2\frac{1}{cos^2t}dt$
$I_{2}=\frac{5}{4}\int\limits_{0}^{\frac{\pi}{4}}\frac{\frac{1}{cos^2t}}{tan^2t+1}dt$=$\frac{5}{4}t [0-\frac{\pi}{4}]$