1) $I=\int\limits \frac{dx}{x^2+x+1}=\int\limits \frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}$đặt $x+\frac{1}{2}=\frac{\sqrt{3} }{2}\tan t\Rightarrow dx=\frac{\sqrt{3} }{2}\frac{1}{\cos^2 t}dt=\frac{\sqrt{3} }{2}(1+\tan^2 t)dt$
$\Rightarrow I=\int\limits \frac{\frac{\sqrt{3} }{2}(1+\tan^2 t)dt}{\frac{3}{4}(1+\tan^2 t)}=\frac{2\sqrt{3} }{3}t+C=\frac{2\sqrt{3} }{3}\arctan \frac{x+\frac{1}{2}}{\frac{\sqrt{3} }{2}}+C$
2) $I=\int\limits (2x-3)e^{2x}dx$
đặt $2x=t\Rightarrow 2dx=dt$
$\Rightarrow I=\int\limits (t-3)e^{t}\frac{dt}{2}=\frac{1}{2}\int\limits(t.e^t-3.e^t)dt=\frac{1}{2}\int\limits t.e^tdt-\frac{3}{2}\int\limits e^tdt$ (sử dụng nguyên hàm từng phần ở $\frac{1}{2}\int\limits t.e^tdt$ tự làm tiếp nhé!)