[tex]\displaystyle I = \int_{0}^3\sqrt{x^2+16} = \int_{0}^3 \dfrac{x^2+16}{\sqrt{x^2+16}}\, dx =\\ \\ =\int_{0}^3 \dfrac{x\cdot x}{\sqrt{x^2+16}}\, dx +\int_{0}^3 \dfrac{16}{\sqrt{x^2+4^2}}\, dx = \\ \\ = \int_{0}^{3}\left(\sqrt{x^2+16}\right)'\cdot x\, dx +16\ln\left(x+\sqrt{x^2+16}\right)\Bigg|_{0}^3 = \\ \\ = \left(\sqrt{x^2+16}\cdot x\right)\Bigg|_{0}^3-\int_{0}^3\sqrt{x^2+16}\cdot x'\, dx +16\ln8-16\ln 4= \\ \\ =5\cdot 3-0-I+16\ln 2[/tex]
[tex]\\[/tex]
[tex]I+I = 15+16\ln 2\\ \\ 2I= 15+16\arctan\dfrac{3}{4}\\ \\ I = \dfrac{15}{2}+8\ln 2[/tex]
