$I=\int\limits_{0}^{3}\frac{2x-1}{x^2+4x+5}dx=\int\limits_{0}^{3}\frac{2x+4}{x^2+4x+5}dx-\int\limits_{0}^{3}\frac{5}{x^2+4x+5}dx=I1-I2$Đặt $t=x^2+4x+5\Rightarrow dt=(2x+4)dx$
Đổi cận: $x=0\Rightarrow t=5,x=3\Rightarrow t=26$
$I1=\int\limits_{5}^{26}\frac{1}{t}dt=ln(t)|^{26}_{5}=ln\frac{26}{5}$
$I2=5\int\limits_{0}^{3}\frac{1}{x^2+4x+5}dx=5\int\limits_{0}^{3}\frac{1}{(x+2)^2+1}dx$
Đặt $tan(u)=(x+2)\Rightarrow \frac{1}{cos^2(u)}du=dx$
Đổi cận: $x=0\Rightarrow u=arctan(2),x=3\Rightarrow u=arctan(5) $
$I2=5\int\limits_{arctan(2)}^{arctan(5)}(\frac{1}{tan^2(u)+1})(\frac{1}{cos^2(u)})du=5.\int\limits_{arctan(2)}^{arctan(5)}1.du=5[arctan(5)-arctan(2)]$
Vậy $I=I1-I2=ln(\frac{26}{5})-5[arctan(5)-arctan(2)]$(Thử lại kết quả bằng casio nhớ chuyển sang radian)