I started from the following problem, known as Jarnik's theorem: Let $f,g:J \longrightarrow\mathbb{R}$ two functions with the following properties:
$f$ and $g$ have antiderivatives;
$g(x)\neq 0,\forall x\in J$.
Then the function $\frac{f}{g}$ has the intermediate value property on $J$. Clearly, if $f$ and $g$ have antiderivatives, then $fg$ does not necessary need to have antiderivatives. Many theorems, such as Wilkosz's theorem, give necessary conditions such that the product of two functions to have antiderivatives. My problem arose when I wanted to find two functions $f$ and $g$, both defined on an interval $J$ with real values, such that $\frac{f}{g}$ does not have an antiderivative. I found the following example: $g:[-1,1]\longrightarrow\mathbb{R}, g(0)=1/2$ and for $x\neq 0$ the graph of $g$ is built of equal parts of isosceles triangles of height $1$ on the segments $[\frac{1}{n+1},\frac{1}{n}],[-\frac{1}{n},-\frac{1}{n+1}], n\in \mathbb{N}$ (positive integer). How can one prove that this function has antiderivatives? On the other side, for $a>0$, the function $f:[-1,1]\longrightarrow\mathbb{R}$, $f(x)=\left\{\begin{matrix} \frac{a}{a/2+g(x)},x\in [-1,1],x\neq 0\\\frac{2a}{a+1},x=0 \end{matrix}\right. $
How can one prove that $f$ doesn't have antiderivatives, but is the quotient of two functions with antiderivatives?