Gợi ý $I=\dfrac{1}{2}\int \dfrac{2e^x .e^x dx}{3+\sqrt{2e^x -1}}$ đặt $\sqrt{2e^x-1}=t \Rightarrow 2e^x -1 = t^2 \Rightarrow e^x dx =tdt$$I=\dfrac{1}{2} \int \dfrac{t^2 +1}{3+t}.tdt =\dfrac{1}{2} \int \bigg (t^2-3t+10 -\dfrac{30}{t+3} \bigg )dt$ dễ rồi nhé
Gợi ý $I=\dfrac{1}{2}\int \dfrac{2e^x .e^x dx}{3+\sqrt{2e^x -1}}$ đặt $\sqrt{2e^x-1}=t \Rightarrow 2e^x -1 = t^2 \Rightarrow e^x dx =tdt$$I=\dfrac{1}{2} \int \dfrac{t^2 +1}{3+t}dt =\dfrac{1}{2} \int \bigg (t-3 +\dfrac{10}{t+3} \bigg )dt$ dễ rồi nhé
Gợi ý $I=\dfrac{1}{2}\int \dfrac{2e^x .e^x dx}{3+\sqrt{2e^x -1}}$ đặt $\sqrt{2e^x-1}=t \Rightarrow 2e^x -1 = t^2 \Rightarrow e^x dx =tdt$$I=\dfrac{1}{2} \int \dfrac{t^2 +1}{3+t}
.tdt =\dfrac{1}{2} \int \bigg (t
^2-3
t+
10 -\dfrac{
30}{t+3} \bigg )dt$ dễ rồi nhé