TITLE: Why isn't there a good product formula for antiderivatives?
QUESTION [8 upvotes]: Computing antiderivatives is more challenging than computing derivatives, in part due to the lack of a ``product formula''; namely, while $(fg)'$ can be expressed in terms of $f,f',g,g'$, there seems to be no way to express $\int fg$ in terms of $f, \int f, g \int g$ and related quantities. Is there any intuitive reason or heuristic explanation for why no such formula exists? I'm looking for a non-rigorous explanation which can be understood by first-year calculus students.

REPLY [10 votes]: Expanding on the idea of @Bill Dubuque, let us hypothesize that we have a differentiable function $p(x_1,x_2,x_3,x_4)$ such that $\int fg=p(f,g,\int f,\int g)$ for all differentiable functions $f$ and $g$. (Assuming $f$ and $g$ are merely continuous is not good enough, since then the right-hand side may not be differentiable, whereas the left-hand side must be.)
More precisely, we assume that whenever $F$ and $G$ are twice differentiable, we have
  $$
  \frac d{dx}p(F'(x),G'(x),F(x),G(x)) = F'(x)G'(x).
  $$
Let us take $F(x)=ux^2+ax+c$ and $G(x)=vx^2+bx+d$. Using the multi-variable chain rule, we get
  \begin{align*}
  ab &= F'(0)G'(0) = \frac d{dx}p(F'(x),G'(x),F(x),G(x))\bigg|_{x=0}\\
  &= 2up_1(a,b,c,d) + 2vp_2(a,b,c,d) + ap_3(a,b,c,d) + bp_4(a,b,c,d).
  \end{align*}
(Here, $p_j=\partial p/\partial x_j$.) Since the left-hand side does not depend on $u$ and $v$, we must have $p_1=p_2=0$, for all $a,b,c,d$. But this implies that the function $p$ depends only on $x_3$ and $x_4$. So the above reduces to
  $$
  ab = ap_3(c,d) + bp_4(c,d).
  $$
Taking $a=1$ and $b=0$ gives $p_3=0$, for all $c,d$. Taking $a=0$ and $b=1$ gives $p_4=0$, for all $c,d$. Finally, then, we have that the function $p$ is a constant function. Since a constant function cannot satisfy our hypothesis, this shows that there can be no such function.