Least integral value..

First of all to simplify the expansion replace $x-1$ by $t$ so that $t>0\forall x >0$ . $f(t+1) = t^{2} + 3{t} + t + \dfrac{1}t^{5}$ We see that the given function is integer only for $t=1$ . Hence we substitute $t=1$ and we get:- $f(2) = 1+3+1+1$ $f(2) = 6$ $f(2)=6$ is the only integral solution.

We can simplify it to $(x-1)^{2}+3(x-1)+1+\frac{1}{(x-1)^{5}}$ which is $(x-1)^{2}+(x-1)+(x-1)+(x-1)+1+\frac{1}{(x-1)^{5}}$ Now, since (x-1)>0 , therefor we can apply the AM -GM>0 inequality , which gives $(x-1)^{ 2 }+(x-1)+(x-1)+(x-1)+1+\frac { 1 }{ (x-1)^{ 5 } } \ge 6\times 1$ Thus the least integral value is 6