Answer
$$\int^1_02\sqrt{x^2+1}dx=\sqrt2+\ln(\sqrt2+1)$$
Work Step by Step
$$A=\int^1_02\sqrt{x^2+1}dx$$
Use Formula 35, which states that
$$\int\sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})+C $$
for $a=1$:
$$A=2\Big[\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\ln(x+\sqrt{x^2+1})\Big]^1_0$$ $$A=x\sqrt{x^2+1}+\ln(x+\sqrt{x^2+1})\Big]^1_0$$ $$A=\Big(1\sqrt{1^2+1}+\ln(1+\sqrt{1^2+1})\Big)-\Big(0\sqrt{0^2+1}+\ln(0+\sqrt{0^2+1})\Big)$$ $$A=(\sqrt2+\ln(\sqrt2+1))-\ln1$$ $$A=\sqrt2+\ln(\sqrt2+1)$$
