Step 1 :The line integral of a scalar function along a curve C is given by the formula: \(\int_{C} f(x,y) ds\), where \(f(x,y)\) is the scalar function and \(ds\) is the differential arc length along the curve C.
Step 2 :In this case, the scalar function is \(f(x,y) = \frac{x}{x^{2}+y^{2}}\) and the curve C is the line segment from \((1,1)\) to \((34,34)\).
Step 3 :Since the curve C is a straight line, we can parameterize it as follows: \(x = t\), \(y = t\) for \(1 \leq t \leq 34\).
Step 4 :Then, the differential arc length \(ds\) can be calculated as \(\sqrt{dx^{2} + dy^{2}} = \sqrt{dt^{2} + dt^{2}} = \sqrt{2} dt\).
Step 5 :Substituting these into the integral, we get: \(\int_{1}^{34} \frac{t}{t^{2}+t^{2}} \sqrt{2} dt\).
Step 6 :This integral can be simplified and then evaluated using standard techniques of calculus.
Step 7 :Let \(t = t\) and \(f = \frac{\sqrt{2}}{2t}\), the integral evaluates to \(\sqrt{2}\log(34)/2\).
Step 8 :Final Answer: \(\boxed{\frac{\sqrt{2}}{2} \log(34)}\)