Cái này hơi dài nhưng cũng không biết còn cách khác không? Bạn tham khảo tạm nhé.
$I=\int\limits_{1}^{2}\frac{2-\sqrt{4-x^2}}{3x^4}dx$.
Đặt $x=2\sin t\Rightarrow dx=2\cos tdt$
Đổi cận $x=1\Rightarrow t=\frac{\pi}{6}, x=2\Rightarrow t=\frac{\pi}{2}$
$\Rightarrow I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{(2-\sqrt{4-4\sin^2t})2\cos tdt}{3.16\sin^4t}=\frac{1}{12}\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{(1-\cos t)\cos t}{\sin^4t}dt$
$=\frac{1}{12}\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\frac{\cos t}{\sin^4t}-\frac{\cos^2t}{\sin^4t})dt=\frac{1}{12}(\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{\cos t}{\sin^4t}dt-\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{\cos^2t}{\sin^4t}dt)=\frac{1}{12}(\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{d(\sin t)}{\sin^4t}+\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2td(\cot t))=\frac{1}{12}(\frac{-1}{3\sin^3 t}+\frac{1}{3}\cot^3t)|^{\frac{\pi}{2}}_{\frac{\pi}{6}}=\frac{1}{12}(\frac{7}{3}-\sqrt{3})$