That looks easy

Make the substitution $x=t^2+1$ $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ $\Rightarrow\sqrt{(t^2+1)+3-4\sqrt{(t^2+1)-1}}+\sqrt{(t^2+1)+8-6\sqrt{(t^2+1)-1}}=1$ $\Rightarrow\sqrt{t^2+1+3-4|t|}+\sqrt{t^2+1+8-6|t|}=1$ $\Rightarrow\sqrt{|t|^2-4|t|+4}+\sqrt{|t|^2-6|t|+9}=1$ Make the substituion $|t|=a$ $\Rightarrow\sqrt{(a-2)^2}+\sqrt{(a-3)^2}=1$ $\Rightarrow |a-2|+|a-3|=1$ $\Rightarrow |3-a|+|a-2|=1$ We can see $(3-a)+(a-2)=1$ And we know that $|a|+|b|\geq|a+b|$ And equality holds when $a$ and $b$ have like sign. So we have, $(3-a)(a-2)\geq 0$ $\Rightarrow (a-3)(a-2)\leq 0$ $\Rightarrow a\in [2,3]$ Reverting back to $t$ $\Rightarrow t\in [2,3]\cup [-3,-2]$ Since $t=\sqrt{a-1}$ , we have $\boxed{x\in [5,10]}$

Let's work the equation: $\sqrt{x+3-4\sqrt {x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ $\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1$ $\sqrt{\sqrt{(x-1)^2}-2*2\sqrt{x-1}+2^2}+\sqrt{\sqrt{(x-1)^2}-2*3\sqrt{x-1}+3^2}=1$ $\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1$ $|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$

Let $\sqrt{x-1}=u$ , where $u≥0$ . Then the equation looks like this:

$|u-2|+|u-3|=1$

There are three cases in terms of $u$ :

$Case$ $1: u\in[0; 2)$ In this case we don't include negative values for $u$ because it is defined only for non-negative values. Here both values in the absolute value bars are negative and therefore out of the bars they are with a minus sign: $-(u-2)-(u-3)=1$ $-2u+5=1$ $2u=4$ $u=2$ which is not in the interval and therefore is not a solution.

$Case$ $2: u\in[2; 3]$ . Here only the value in the second pair of absolute bars is negative so only it is with a minus sign while the other one is with a plus sign: $u-2-(u-3)=1$ $u-2-u+3=1$ This is solved to $1=1$ which is a true equality and so all values of the interval are solutions $\Rightarrow u\in[2; 3]$ .

$Case$ $3: u\in(3; +∞]$ . In this case both values in the bars are positive and so both of them are with a plus sign when absolute value bars are removed: $u-2+u-3=1$ $2u=6$ $u=3$ which is not in the interval for this case and therefore in this case it is not a solution.

To sum up: From the first and the third case we don't have solutions and for the second one we have $u\in[2; 3]$ and so all the solutions are from the second case. This can be expressed as a double inequality: $2≤u≤3$ When we substitute in $u=\sqrt{x-1}$ we get: $2≤\sqrt{x-1}≤3$ To solve this for $x$ we need to do the same operations with all the three values ( $2, \sqrt{x-1}, 3$ in the beginning). $2≤\sqrt{x-1}≤3 \rightarrow\textit{Raise to power 2}$ $4≤x-1≤9 \rightarrow \textit{Add 1}$ $5≤x≤10$ Therefore we can conclude that all solutions to the equation, which are integers, are $5, 6, 7, 8, 9$ and $10$ and they sum up to $\fbox{45}$ .

cheated...... solving numerically. i'm an engineer.

Let √(x-1) = a, x = a2 + 1, √(x+3-4√(x-1)) + √(x+8-6√(x-1)) = √(a^2+4-4a)+√(a^2+9-6a) = 1 √(a-2)^2 +√( a-3)^2 = 1, or |a – 2| + |a – 3| = 1 a) let a≥3, then 2a = 6, or a=3 b) let 2≤a<3, then 1=1, always true. So √(x-1) = 3, then x=10, 2≤√(x-1)<3, 5≤x<10. And the integer solutions are; 5,6,7,8,9,10. The sum = 45