With calculus, define the function $$f(a,b)=\frac{a}{b}$$ Then the change resulting from the addition of a small $\epsilon$ will be approximately $$df\approx f_ada+f_bdb={1\over b}da-{a\over b^2}db={\epsilon\over b^2}(b-a)$$ This change will be positive if $\epsilon$ and $b-a$ have the same sign.
So the answer is no. For example, take $\epsilon=0.5$ and $a=2,b=1$. Then the original fraction is $2$ and the new one is $$\frac{2.5}{1.5}=\frac{5}{3}<2$$
In the integral $\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$ , the numbers $5$, $4$ and
$3$ are very carefully chosen such that the integrand can be simplified. It is not possible to solve the integral with any $a$, $b$ and $c$ ; instead of some specific carefully choosen $a$, $b$ and $c$.
Now, $$\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}dx = \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{(\sin 3x)(1-2\cos 3x)}dx$$
I have multiplied numerator and denominator by $\sin 3x$ to remove the coefficient "2" of $\cos 3x$ since $2\sin 3x\cos 3x$ would yield $\sin 6x$ (in general, $p$ should be equal to $c$ to remove that "2"). I am removing this "2" to apply the formula of $\sin A - \sin B$, in the hope that, doing similar thing (i.e. applying formula of $\cos A + \cos B$) in numerator would lead to cancellation of some common terms from numerator and denominator (that's exactly what will happen if you notice the solution further).
The cancellation is because of the choice of appropriate angles ($5x,4x$ and $3x$) of sine and cosine. This cancellation would not be possible if the angles are randomly chosen. What I mean to say is that besides having $p$ to be equal to $c$ , $a$ and $b$ should be well chosen so that the integrand can be simplified.
$$\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}dx= \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{\sin 3x-2\sin 3x.\cos 3x}dx$$
$$= \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{\sin 3x-\sin 6x}dx$$
$$= \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{2\cos \frac{9x}{2}.\sin\frac{-3x}{2}}dx$$
$$= \int \dfrac{(\sin 3x)(\require{cancel}\cancel{2} \cancel{\cos \frac{9x}{2}}.\cos \frac {x}{2})}{\cancel{2}\cancel{\cos \frac{9x}{2}}.\sin\frac{-3x}{2}}dx$$
$$= \int \dfrac{(2\cancel{\sin \frac{3x}{2}}.\cos \frac{3x}{2})(\cos \frac{x}{2})}{(-\cancel{\sin \frac{3x}{2}})}dx$$
$$= -\int (2\cos \frac{3x}{2}.\cos \frac{x}{2})dx$$
$$= -\int (\cos 2x+\cos x)dx$$
$$= \int (-\cos 2x-\cos x)dx$$
$$= -\dfrac {\sin 2x}{2} - \sin x + c$$
Best Answer
You are right. There is a problem with this. Integrating, $$ \int\frac{du}{u^2+3} = \frac{1}{\sqrt{3}}\arctan\left(\frac{u}{\sqrt{3}}\right) =\frac{1}{\sqrt{3}}\arctan\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right) =: F_1(x) $$ we get this:
Whereas, a good antiderivative is $$ F_2(x) := \frac{1}{\sqrt{3}}\left(\arctan\frac{(2x+1)}{\sqrt{3}}+ \arctan\frac{(2x-1)}{\sqrt{3}}\right) $$ and looks like this
On any interval $[a,b]$ with $0 \notin [a,b]$ we have $F_1-F_2$ is constant on $[a,b]$. If we want to use $F_1$ to evaluate an integral $\int_a^b$ with $a < 0 < b$, we have to do it in two parts: $$ \int_a^0 + \int_0^b = (F_1(0^-)-F_1(a))+(F_1(b)-F_1(0^+)). $$ Of course this is not the only example of this phenomenon. So when evaluating an integral using the Fundamental Theorem, we should always check for jumps in our antiderivative and act accordingly.