Use Green's theorem to evaluate $\int_{C} F \cdot d \mathbf{r}$. (Check the orientation of the curve before applying the theorem.) $F(x, y)=\left\langle\sqrt{x^{2}+2}, \tan ^{-1}(x)\right\rangle, \quad C$ is the triangle from $(0,0)$ to $(1,1)$ to $(0,1)$ to $(0,0)$

Step 1 ：Given the vector field \(F(x, y)=\left\langle\sqrt{x^{2}+2}, \tan ^{-1}(x)\right\rangle\), and the curve \(C\) is the triangle from \((0,0)\) to \((1,1)\) to \((0,1)\) to \((0,0)\)

Step 2 ：We need to evaluate the line integral \(\int_{C} F \cdot d \mathbf{r}\) using Green's theorem. Green's theorem states that for a simply connected region \(D\) bounded by a piecewise smooth simple closed curve \(C\), if \(F\) is a vector field whose components have continuous first partial derivatives on an open region containing \(D\), then the line integral of \(F\) over the boundary of \(D\) is equal to the double integral of the divergence of \(F\) over \(D\).

Step 3 ：The divergence of \(F\) is given by the partial derivative of the first component of \(F\) with respect to \(x\) plus the partial derivative of the second component of \(F\) with respect to \(y\). In this case, the divergence of \(F\) is \(x/\sqrt{x^{2} + 2}\).

Step 4 ：We then calculate the double integral of the divergence of \(F\) over the region \(D\), which is the triangle from \((0,0)\) to \((1,1)\) to \((0,1)\) to \((0,0)\). The result of the double integral is \(-\sqrt{2} + \sqrt{3}\).

Step 5 ：Finally, we check the orientation of the curve. If the curve is oriented counter-clockwise, then the result of the double integral is the answer. If the curve is oriented clockwise, then the result of the double integral is the negative of the answer. In this case, the curve is oriented counter-clockwise, so the result of the double integral is the answer.

Step 6 ：Thus, the value of the line integral \(\int_{C} F \cdot d \mathbf{r}\) is \(\boxed{-\sqrt{2} + \sqrt{3}}\).