Answer
$F'(x) = -\sqrt{x+sin~x}$
Work Step by Step
The function $f(t) = \sqrt{t+sin~t}$ is continuous on the interval $(1,\infty)$
Let $G(x) = \int_{1}^{x}f(t)~dt$
According to the Fundamental Theorem of Calculus (Part 1):
$G'(x) = f(x)$
Therefore:
$G'(x) = \sqrt{x+sin~x}$
Note that $F(x) = \int_{x}^{1}f(t)~dt$
Then $F'(x) = -G'(x)$
Therefore:
$F'(x) = -\sqrt{x+sin~x}$
