Can I reduce fraction under integral? Can I perform such kind of fraction reduction under integral?
$$g(x)=\int \frac{d f(x)} {dx} dx = \int d f(x) = f(x) +C $$
 A: The answer is yes. In order to avoid confusing fractions and differentials everywhere, I will write this in purely operator form to emphasize how simple this is. We can denote differentiation as a map between sets of differentiable/integrable functions, i.e
$$\mathcal{D}:\mathscr{D}\to\mathscr{I}~~,~~\mathcal{D}:f\mapsto \mathcal{D}(f)$$
Integration can be written similarly,
$$\mathcal{I}:\mathscr{I}\to\mathscr{D}~~,~~\mathcal{I}:f\mapsto\mathcal{I}(f)$$
Integration is defined to be an operation such that
$$(\mathcal{I}\circ\mathcal{D})=(\mathcal{D}\circ\mathcal{I})=\operatorname{id}$$
$$(\mathcal{I}\circ\mathcal{D})(f)=(\mathcal{D}\circ\mathcal{I})(f)=f$$
Where equality is taken to be up to a constant difference. This means, in words, that taking the derivative of the primitive of a function or taking the primitive of the derivative of a function will give you back the function. If we use more classical notation, this is
$$\int\frac{\mathrm{d}f}{\mathrm{d}x}\mathrm{d}x=\frac{\mathrm{d}}{\mathrm{d}x}\int f~\mathrm{d}x=f$$
A: Unless you want to work in nonstandar analysis, the Leibniz notation $\frac{df(x)}{dx}$ is not a fraction. It means the derivative of the function $f$; in Lagrange's notation, it is the same as $f'(x)$.
The identity $\int f'(x)dx=f(x)+C$ is due to the (trivial) fact that $f$ is an antiderivative of $f$ (on some interval).
The identity $\int f'(x)dx=\int df(x)$ could be understood as change of variables. If there is anything to justify, you must define what the right-hand side means first.
A: Your outcome is correct but it's not a matter of reducing the fraction.  This is more a case where the notation is chosen to line up with the result, not the other way around  Remember that $$\frac{df(x)}{dx} = \text{ the derivative of $f$ with respect to $x$}$$ and this is not a fraction.  Also, $$\int h(x)\,dx = \text{the antiderivative of $h$ with respect to $x$}.$$  So, combining these: $$\int \frac{df(x)}{dx}\,dx = \text{the antiderivative of $\frac{df(x)}{dx}$ with respect to $x$},$$ and the antiderivative of a derivative is the original function plus $C$, so we have $$\int \frac{df(x)}{dx}\,dx = f(x) + C$$
