Đặt $t=\sqrt{1+x^2}\Rightarrow t^2=1+x^2\Leftrightarrow t^2-1=x^2\Rightarrow tdt=xdx. $ Đổi cận:...khi đó $I=\int\limits_{1}^{\frac{\sqrt{9+\pi^2}}{3}} (t^2-1)t.tdt=\int\limits_{1}^{\frac{\sqrt{9+\pi^2}}{3}}(t^4-t^2)dt=(\frac{t^5}{5}-\frac{t^3}{3})|_1^\frac{\sqrt{9+\pi^2}}{3}=..$