f(x) = x⁵ + 5x – 1
f(1) = 5  ⇒  f ^–1(5) = 1
f(2) = 41  ⇒  f ^–1(41) = 2
Put  f ^–1(x) = t                 
∴ $\mathrm{I} =\underset{1}{\overset{2}{\int }}\frac{\mathrm{f}\text{'}\left(\mathrm{t}\right)}{{\mathrm{t}}^5+5\mathrm{t}}\mathrm{dt}=\underset{1}{\overset{2}{\int }}\frac{5{\mathrm{t}}^4+5}{{\mathrm{t}}^5+5\mathrm{t}}\mathrm{dt}={\ln ({\mathrm{t}}^5+5\mathrm{t})]}_1^2 =$  ln 42 – ln 6 = ln 7.