$\int\limits_{1}^{\sqrt{3}}\frac{\sqrt{9+3x^{2}}dx}{x^{2}}$
Không khó lắm đâu 

$I=-\dfrac{1}{9} \int_1^{\sqrt 3} x^2.\dfrac{\sqrt{9+3x^2}}{x} . \dfrac{-9 dx}{x^3}$

Đặt $\dfrac{\sqrt{9+3x^2}}{x}=t\Rightarrow \dfrac{9+3x^2}{x^2}=t^2 \Rightarrow \dfrac{9}{x^2}+3 =t^2$

$\Rightarrow -\dfrac{9dx}{x^3}=tdt$

Vậy $I=\dfrac{1}{9} \int \limits_{\sqrt 6}^{2\sqrt 3} \dfrac{9}{t^2-3}. t^2 dt = \dfrac{1}{9} \int \limits_{\sqrt 6}^{2\sqrt 3} \bigg (9+\dfrac{27}{t^2-3} \bigg )dt$

Giờ thì ok rồi nhé, cái $\int \dfrac{1}{t^2-3}dt =\dfrac{1}{2\sqrt 3}\int \bigg (\dfrac{1}{t-\sqrt 3}-\dfrac{1}{t+\sqrt 3} \bigg )dt$
ý chú là sao :3 –  Dép Lê Con Nhà Quê 29-12-13 06:20 PM
ông anh nhin thế mà nguy hiểm vãi –  Phạm Anh Tuấn 29-12-13 02:03 AM
k co j b, mih làm ẩu quá nên làm b mất tjan –  Dép Lê Con Nhà Quê 28-12-13 08:24 PM
ah uh tôi lại đi lấy nguyeen hàm, để tý sửa lại cho –  Dép Lê Con Nhà Quê 28-12-13 07:41 PM

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