Text Compute ∫022xx2+1 dx\displaystyle\int_0^2 2x\sqrt{x^{2}+1}\,dx∫022xx2+1dx by substitution. Solution Let u=x2+1u=x^{2}+1u=x2+1, du=2x dxdu=2x\,dxdu=2xdx; x=0⇒u=1x=0\Rightarrow u=1x=0⇒u=1, x=2⇒u=5x=2\Rightarrow u=5x=2⇒u=5. ∫15u du=[23u3/2]15=23(55−1)≈6.787\displaystyle\int_1^5 \sqrt{u}\,du=\Big[\tfrac23 u^{3/2}\Big]_1^5=\tfrac23(5\sqrt5-1)\approx6.787∫15udu=[32u3/2]15=32(55−1)≈6.787. I=23(55−1)≈6.787\boxed{I=\tfrac23(5\sqrt5-1)\approx6.787}I=32(55−1)≈6.787