Given integral equation

$\int_0^x \sqrt{1-\left(y'(t)\right)^2}\,dt=\int_0^x y(t)\,dt, \qquad 0\le x\le 3$

with $y(x)\ge 0, \qquad y(0)=0.$

We need to find the value of $y''+y+1$ at $x=2$.

Differentiate both sides with respect to $x$

By the Fundamental Theorem of Calculus, $\sqrt{1-(y'(x))^2}=y(x).$

Since $y\ge 0$, this is consistent with the square root being nonnegative.

Square both sides

$1-(y')^2=y^2$ so $(y')^2+y^2=1.$

This is the key relation.

Use the initial condition to determine the solution branch

At $x=0$, we have $y(0)=0$. Then from $(y')^2+y^2=1,$ $(y'(0))^2=1 \implies y'(0)=\pm 1.$

Also, from $\sqrt{1-(y')^2}=y,$ near $x=0$, $y\ge 0$.

Differentiate $(y')^2+y^2=1$ to get $2y'y''+2yy'=0$ $\Rightarrow y'(y''+y)=0.$

For nontrivial intervals where $y'\ne 0$, $y''+y=0.$

With $y(0)=0$ and $y'(0)=\pm1$, the solutions are $y=\sin x \quad \text{or} \quad y=-\sin x.$

Since $y(x)\ge 0$ for $0\le x\le 3$, and $\sin x>0$ on $(0,3)$, the valid solution is $y(x)=\sin x.$

Compute the required expression

For $y=\sin x,$ $y''=-\sin x=-y.$

Therefore, $y''+y+1=-y+y+1=1.$

So at $x=2$, $y''(2)+y(2)+1=1.$

Check options

The correct option is: $\boxed{\text{D: }1}$