$\int\limits_{2}^{4} \frac{3x-1}{x^{2}-4x+8} dx$=$\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$$I_{2}=\frac{5}{4}\int\limits_{0}^{\frac{\pi}{4}}\frac{\frac{1}{cos^2t}}{tan^2t+1}dt$=$\frac{5}{4}t [0-\frac{\pi}{4}]$
$\int\limits_{2}^{4} \frac{3x-1}{x^{2}-4x+8} dx$=$\frac{1}{3}\int\limits_{2}^{4}\frac{2x-4+1}{x^2-4x+8}dx$=$\frac{1}{3}\int\limits_{2}^{4}\frac{d(x^2-4x+8)}{x^2-4x+8}(I)+\frac{1}{3}\int\limits_{2}^{4}\frac{dx}{x^2-4x+8}(I_{2})$$=\frac{1}{3}ln\left| {x^2-4x+8} \right| [2-4]+I_{2}$$I_{2}$ $=\frac{1}{3}\int\limits_{2}^{4}\frac{dx}{(x-2)^2+4}$ Đặt $x-2=2tant$ $dx=2\frac{1}{cos^2t}dt$$I_{2}=\frac{1}{6}\int\limits_{0}^{\frac{\pi}{4}}\frac{\frac{1}{cos^2t}}{tan^2t+1}dt$=$\frac{1}{6}t [0-\frac{\pi}{4}]$
$\int\limits_{2}^{4} \frac{3x-1}{x^{2}-4x+8} dx$=$\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}]$