The arc length is given by the formula ∫√(1+x-2/3)dx with limits x=1 and x=8. The integration is rather tricky. Here's how I went about it:
Begin with the substitution u=x-1/3. Then x=u-3 so that dx=-3u-4du. The integral becomes ∫-3u-4√(1+u2)du with limits u=1 and u=1/2.Note that 1/2 is the upper limit.
Now use the substitution u=tanθ. Then du=sec2θdθ. The integral becomes ∫-3sec3θ⋅(tanθ)-4dθ with limits θ=π/4 and θ=tan-1(1/2). Again, tan-1(1/2) is the upper limit, despite it being less than π/4.
Rewrite the last integral using sines and cosines and you'll get ∫-3cosθ(sinθ)-4dθ, which we can finally integrate to get (sinθ)-3. Evaluating at the limits on θ gives the arc length as 5√5 - 2√2.