Find $\int_{C} x d y+y d x$, where $C$ is the arc of the curve $y=e^{x}$ for $0 \leq x \leq 6$.

Step 1 ：Given the integral \(\int_{C} x d y+y d x\), where \(C\) is the arc of the curve \(y=e^{x}\) for \(0 \leq x \leq 6\).

Step 2 ：This is a line integral problem. The line integral of a scalar field along a curve C is defined as the integral of the scalar field along the path defined by the curve. In this case, the scalar field is given by the function \(f(x, y) = x dy + y dx\) and the curve C is defined by the function \(y = e^x\) for \(0 \leq x \leq 6\).

Step 3 ：The line integral of a scalar field along a curve C is given by the formula: \(\int_{C} f(x, y) ds = \int_{a}^{b} f(x(t), y(t)) ||r'(t)|| dt\), where \(r(t) = (x(t), y(t))\) is a parameterization of the curve C and \(||r'(t)||\) is the magnitude of the derivative of \(r(t)\).

Step 4 ：In this case, we can parameterize the curve C by the function \(r(t) = (t, e^t)\) for \(0 \leq t \leq 6\). Then, the derivative of \(r(t)\) is \(r'(t) = (1, e^t)\) and the magnitude of \(r'(t)\) is \(||r'(t)|| = \sqrt{1 + e^{2t}}\).

Step 5 ：Then, we can substitute these values into the formula for the line integral to find the value of the integral. We have \(t = t\), \(x = t\), \(y = e^t\), \(dx = 1\), \(dy = e^t\), \(f = t*e^t + e^t\), \(ds = \sqrt{e^{2t} + 1}\), and the integral is \(\int_{0}^{6} (t + 1)*\sqrt{e^{2t} + 1}*e^t dt\).

Step 6 ：The numerical value of the integral is approximately 528964.742758784.

Step 7 ：Final Answer: The value of the line integral \(\int_{C} x d y+y d x\), where \(C\) is the arc of the curve \(y=e^{x}\) for \(0 \leq x \leq 6\) is approximately \(\boxed{528964.742758784}\).