Không khó lắm đâu $I=-\dfrac{1}{9} \int_1^{\sqrt 3} \dfrac{\sqrt{9+3x^2}}{x} . \dfrac{-9 dx}{x}$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$$\Rightarrow -\dfrac{9dx}{x}=2tdt$Vậy $I=\dfrac{1}{9} \int_{\sqrt 6}^{2\sqrt 3} 2t^2 dt =\dfrac{2}{27}t^3 +C$ tự thay cận nhé
Không khó lắm đâu $I=-\dfrac{1}{9} \int_1^{\sqrt 3}
x^2.\dfrac{\sqrt{9+3x^2}}{x} . \dfrac{-9 dx}{x
^3}$Đặt $\dfrac{\sqrt{9+3x^2}}{x}=t\Rightarrow \dfrac{9+3x^2}{x^2}=t^2 \Rightarrow \dfrac{9}{x^2}+3 =t^2$$\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 th
ì ok rồi nhé
, cái $\int \dfrac{1}{t^2-3}dt =\dfrac{1}{2\sqrt 3}\int \bigg (\dfrac{1}{t-\sqrt 3}-\dfrac{1}{t+\sqrt 3} \bigg )dt$