Có: \(I=\int\limits^{ln3}_0\sqrt{e^x-1}e^xdx\)
Đặt \(t=\sqrt{e^x-1}\Rightarrow t^2=e^x-1\)
\(\Rightarrow2tdt=e^xdx\)
\(\Rightarrow I=\int\limits^{\sqrt{2}}_02t^2dt\) \(\Rightarrow I=\dfrac{2}{3}t^3|^{\sqrt{2}}_0=\dfrac{4}{3}\sqrt{2}\)
Vậy a=0, b=\(\dfrac{4}{3}\) \(\Rightarrow a+b=0+\dfrac{4}{3}=\dfrac{4}{3}\)