Expected number of comparisons = sum((number of comparison till ith entry) * probability of getting the number at ith entry) + n. probability that $x$ is not in the list.
Probability that $x$ is in the list = $\sum_{i=1}^n \frac{ i} {n.(n+1)} = \frac{1}{2}$
So, probability of $x$ not in the list $= \frac{1}{2}$
Thus, Expected number of comparisons
$= \sum_{i=1}^n \frac{ i} {n.(n+1)} . i + \frac{n}{2}\\= \sum_{i=1}^n \frac{i^2}{n. (n+1)} + \frac{n}{2} \\= \frac{2n+1}{6} + \frac{n}{2} \because \sum_{i=1}^n i^2 = \frac{n.(n+1).(2n+1)}{6} \\= \frac{5n + 1}{6} $
Now, there is a catch here. We have counted one comparison for one element in the linear list. But if we count the loop exit condition and the final found or not check also as in the algorithm give here http://www.programmingsimplified.com/c/source-code/c-program-linear-search, we get $2i + 1$ comparisons for a successful check at $i^{th}$ position and $2n+2$ comparisons if $x$ is not in the list. This would give
$\text{Expected number of comparisons} = \sum_{i=1}^n \frac{ i} {n.(n+1)} . (2i+1) + \frac{2n+2}{2}\\= \sum_{i=1}^n \frac{2i^2 + i}{n. (n+1)} +(n+1) \\= \frac{2n+1}{3} + \frac{1}{2} + (n+1) \\= \frac{5n+4}{3} + \frac{1}{2} = \frac{10n + 11}{6} $