Is it ok to do this change of variable in integration: let $x = x - 1$ In integrals like $\int \sqrt{x-1}\,dx$, is it ok to make  this change of variable in integration: "let $x = x - 1$"?
It looks sketchy — like saying, let 5 = 4.
 A: No, but you can let $u=x-1$. $x$ is never equal to $x-1$.
A: 
Is it ok to do this change of variable in integration: let $x = x - 1$

Certainly you may, if you do so carefully.
$$\begin{align}
g(x)+\int_{a}^{b} f(x) \operatorname d x
 & \;=\; g(x)+\int_{a}^{b} f(x-1) \operatorname d (x-1)
\\[1ex] & \;=\; g(x)+\int_{a+1}^{b+1} f(x-1) \operatorname d x
\end{align}$$
Notice, when we do this we may only so change the variable bound within the integration's scope.
It is probably safer to substitute with a different variable name though. (Less confusing.)
$$\begin{align}
g(x)+\int_{a}^{b} f(x) \operatorname d x
 & \;=\; g(u-1)+\int_{a}^{b} f(u-1) \operatorname d (u-1)
\\[1ex] & \;=\; g(u-1)+\int_{a+1}^{b+1} f(u-1) \operatorname d u
\end{align}$$
Notice: here we can change the variable outside the integration scope.
You might more often see written out as:
$$\begin{align}
\text{ Let } x & = u-1 & \therefore \mathrm d x = \mathrm d u
\\[2ex] g(x)+\int_{a}^{b} f(x) \operatorname d x
 & \;=\; g(u-1)+\int_{a+1}^{b+1} f(u-1) \operatorname d u
\end{align}$$
