## Thomas' Calculus 13th Edition

$$\dfrac{7}{3}$$
We calculate the line integral by using substitution as follows: $$ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt\\ =\sqrt {t^2} dt \\= t \space dt$$ $$\int_C f(x,y) ds=\int_{0}^{\sqrt 3} \sqrt {(1+t^2)} t \space dt \\= (\dfrac{1}{3})[(1+t^2)^{3/2}]_{0}^{\sqrt 3} \\=\dfrac{7}{3}$$