1. Given data

   $f(x)=\int_0^x g(t)\,dt$ and $g$ is a non-zero even function, so $g(-x)=g(x).$

   Also, $f(x+5)=g(x).$

2. Use parity of $g$ to find parity of $f$

   Since $g$ is even, $f(-x)=\int_0^{-x} g(t)\,dt.$

   Put $t=-u$, then $dt=-du$: $f(-x)=\int_0^x g(-u)(-du)= -\int_0^x g(u)\,du = -f(x).$

   Hence, $f$ is an odd function.

3. Relate $g$ and $f$ using the given condition

   Given $f(x+5)=g(x).$

   Replace $x$ by $t-5$: $f(t)=g(t-5).$

   Therefore, $\int_0^x f(t)\,dt = \int_0^x g(t-5)\,dt.$

4. Change variable in the integral

   Let $u=t-5 \implies du=dt.$

   When $t=0$, $u=-5$. When $t=x$, $u=x-5$.

   So, $\int_0^x f(t)\,dt = \int_{-5}^{x-5} g(u)\,du.$

5. Use the fact that $g$ is even

   Since $g$ is even, $\int_{-5}^{x-5} g(u)\,du = \int_0^{x-5} g(u)\,du + \int_{-5}^0 g(u)\,du.$

   Now, $\int_{-5}^0 g(u)\,du = \int_0^5 g(u)\,du$ because $g$ is even.

   Hence,

   $$$$

   This is not yet in the options. So let us use the oddness of $f$ more directly.

6. Better approach using oddness of $f$

   Since $f$ is odd, $f(x+5)=g(x)$ implies $f(-(x+5))=-f(x+5)=-g(x).$

   But $f(-(x+5))=f(-x-5)=g(-x-10)?$ This does not directly help. Instead, integrate the given relation carefully.

7. Express $f(t)$ from the given relation

   From $f(x+5)=g(x),$ set $x=t-5$: $f(t)=g(t-5).$

   Therefore, $\int_0^x f(t)dt = \int_0^x g(t-5)dt = \int_{-5}^{x-5} g(u)du.$

   Now compare with option B: $\int_{x+5}^{5} g(t)dt = -\int_5^{x+5} g(t)dt.$

   Let $t=u+10$? That is not generally equal. So we should derive from the given relation another way.

8. Differentiate the given relation

   Since $f(x)=\int_0^x g(t)dt,$ by FTC, $f'(x)=g(x).$

   Given also, $f(x+5)=g(x).$

   Therefore, $f(x+5)=f'(x).$

   Replace $x$ by $t$: $f(t+5)=f'(t).$

9. Integrate this relation from $0$ to $x$

   $\int_0^x f(t+5)dt = \int_0^x f'(t)dt = f(x)-f(0).$

   Since $f(0)=\int_0^0 g(t)dt=0,$ we get $\int_0^x f(t+5)dt=f(x).$

   Let $u=t+5$, then $\int_5^{x+5} f(u)du=f(x).$

   Since $f(x)=\int_0^x g(t)dt$, this still does not directly match the target. Instead, integrate $f'(t)=f(t+5)$ from $-5$ to $x-5$:

   $\int_{-5}^{x-5} f'(t)dt = \int_{-5}^{x-5} f(t+5)dt.$

   So, $f(x-5)-f(-5)=\int_0^x f(u)du.$

10. Use oddness of $f$ and given relation

Since $f$ is odd, $f(-5)=-f(5).$

Also from $f(x+5)=g(x)$, putting $x=0$ gives $f(5)=g(0).$

And $f(x-5)=g(x-10).$

This again does not seem to simplify to the options directly. So let us derive using the relation $f'(x)=f(x+5)$ in a cleaner way.

11. Observe a key identity

We want $I(x)=\int_0^x f(t)dt.$

Differentiate option B candidate: $J(x)=\int_{x+5}^{5} g(t)dt.$

Then by Leibniz rule, $J'(x)=-g(x+5).$

Since $g$ is even, $J'(x)=-g(x+5)= -g(-(x+5))=-g(-x-5).$

But using $g(y)=f(y+5)$, $g(-x-5)=f(-x)= -f(x),$ because $f$ is odd.

Hence, $J'(x)=-(-f(x))=f(x).$

Also, $J(0)=\int_5^5 g(t)dt=0,$ and $I(0)=\int_0^0 f(t)dt=0.$

Since both $I$ and $J$ have the same derivative and same value at $x=0$, $I(x)=J(x).$

Therefore, $\int_0^x f(t)dt = \int_{x+5}^{5} g(t)dt.$

12. Correct option

$\boxed{\int_0^x f(t)dt = \int_{x+5}^{5} g(t)dt}$

So the correct answer is Option B.