Answer
$\sqrt {2}+\ln(\sqrt2+1)$
Work Step by Step
Apply the formula:
$\int\sqrt{a^2+x^2}dx=\dfrac{x}{2}\sqrt{a^2+x^2}+\dfrac{a^2}{2}\ln(x+\sqrt{a^2+x^2})+C $
Let $I=2[\dfrac{x}{2}\sqrt{x^2+1}+\dfrac{1}{2}\ln(x+\sqrt{x^2+1})]^1_0 \\=[x\sqrt{x^2+1}+\ln(x+\sqrt{x^2+1}) ]^1_0 \\=(\sqrt2+\ln(\sqrt2+1))-\ln1 \\=\sqrt {2}+\ln(\sqrt2+1)$
