How do I justify this technically? I have two differentiable functions, $\mathrm{f}$ and $\mathrm{F}$, with the properties $\mathrm{F}(0)=0$ and
$$\frac{\mathrm{dF}}{\mathrm{d}x} = \mathrm{f}(x)$$
Applying the fundamental theorem of calculus gives:
$$F(x) = \int_0^x \mathrm{f}(t) \ \mathrm{d}t$$
Now, consider the following differential equation, with the initial conditions $y=1$ when $x=0$:
$$\frac{\mathrm{d}y}{\mathrm{d}x}+\mathrm{f}(x)\,y=0$$
This equation is seperable and we obtain:
$$\int\frac{1}{y}\mathrm{d}y =- \int\mathrm{f}(x)\ \mathrm{d}x$$
I need to write this equation in terms of $\mathrm{F}$, so I need to introduce some limits. When $x=0$ we have $y=1$ and when $x=x$ we have $y=y(x)$ and so I feel that I can introduce the limits
$$\int_1^y\frac{1}{\gamma}\mathrm{d}\gamma =- \int_0^x\mathrm{f}(\chi)\ \mathrm{d}\chi $$
How can I technically justify this last step? 
EDIT: I would like to know how to go from the ODE to the final line. One part of this is to justify expressing one integral w.r.t. one variable and with one set of limits in terms of another limit w.r.t. another variable and with different limits. I suspect that a single integration and a cunning substitution would do the trick.
 A: If you divide your differential equation by $y$, it becomes
$$
(\ln(y))^\prime+f=0.
$$
Integrating both sides from $0$ to $x$, this becomes
$$
\int_{0}^x(\ln(y))^\prime(s)\,ds=-\int_0^xf(\chi)\,d\chi.
$$
Now the change-of-variables formula can be applied to the integral on the LHS, giving
$$
\int_{y(0)}^{y(x)}\frac{1}{\gamma}\,d\gamma=-\int_0^x f(\chi)\,d\chi.
$$
edit: Here is a little more explanation of the change-of-variable formula and its application to this problem. Suppose we have a smooth function $u:[a,b]\rightarrow\mathbb{R}$ whose derivative is nowhere zero. The non-vanishing derivative assumption implies that there is an inverse function $u^{-1}$ whose domain is the image of $[a,b]$ under $u$, i.e. $[\min(u(a),u(b)),\max(u(a),u(b))]$. For simplicity, assume $u^\prime$ is positive and that $u^{-1}$ is smooth.
Now consider the integral
$$
I_1(b)=\int_a^bg(s)\,ds.
$$ 
Let
$$
I_2(b)=\int_{u(a)}^{u(b)}g(u^{-1}(r))(u^{-1})^\prime(r)\,dr=\int_{u(a)}^{u(b)}g(u^{-1}(r))\frac{1}{u^\prime(u^{-1}(r))}\,dr.
$$
I claim $I_1(b)=I_2(b)$. In order to see this, first check that $I_1^\prime(b)=I_2^\prime(b)$. By the chain rule and the fundamental theorem of calculus,
$$
I_2^\prime(b)=(u^\prime(b))\left(g(u^{-1}(u(b)))\frac{1}{u^\prime(u^{-1}(u(b)))}\right)=g(b).
$$ 
By the fundamental theorem of calculus (again), 
$$
I_1^\prime(b)=g(b).
$$
So everything checks out with the equality $I_1^\prime(b)=I_2^\prime(b)$. But now we see that 
\begin{align}
(I_1+I_2)^\prime=0,
\end{align}
which implies that $I_1(b)=I_2(b)+c$, where $c$ is a constant that does not depend on $b$. Moreover, $c$ must be zero because $I_1(a)=I_2(a)=0$. This proves my claim that
$$
\int_a^bg(s)\,ds=\int_{u(a)}^{u(b)}g(u^{-1}(r))(u^{-1})^\prime(r)\,dr=\int_{u(a)}^{u(b)}g(u^{-1}(r))\frac{1}{u^\prime(u^{-1}(r))}\,dr.
$$
Here is how this can all be applied to the integral
$$
\int_0^x (\ln(y))^\prime(s)\,ds.
$$
Assume that $y$ has a non-zero derivative on the interval $[0,x]$. In the above formula, set $g=(\ln(y))^\prime$ and $u=y$. We have
$$
\int_0^x (\ln(y))^\prime(s)\,ds=\int_{y(0)}^{y(x)}(\ln y)^\prime(y^{-1}(r))\frac{1}{y^\prime(y^{-1}(r))}\,dr=\int_{y(0)}^{y(x)}\frac{1}{r}\,dr.
$$
