Why isn't there a good product formula for antiderivatives? 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. 
 A: My answer:
Differentiating is somewhat like taking the difference of two things (that are close together).
Integrating is somewhat like taking the sum of many things (that may be fluctuating).
Suppose $f$ and $g$ are functions that we "know" how to differentiate and integrate. That's somewhat analogous to saying we have expressions for
$f_2-f_1$
$g_2-g_1$
$f_1+\cdots+f_n$
$g_1+\cdots+g_n$
Wanting to differentiate or integrate the product $fg$ is somewhat analogous to wanting formulas for $f_2g_2-f_1g_1$ or for $f_1g_1+\cdots+f_ng_n$.
With a difference of two things, you can do algebraic "tricks" such as $f_2g_2-f_1g_1=f_2g_2-f_1g_2+f_1g_2-f_1g_1$.
It's not so easy to find an algebraic "trick" for expressing $f_1g_1+\cdots+f_ng_n$ in terms of $f_1+\cdots+f_n$ and $g_1+\cdots+g_n$.
A: Maybe use the notion of "elementary function". There are elementary functions $A,B$ such that both $\int A$ and $\int B$ are elementary, but $\int(AB)$ is not elementary. So there can be no simple "product rule" for integration, since any simple combinations of elementary functions is again elementary.  
Example: $A(x) = \cos x$, $B(x) = \frac{1}{x}$.  Then $\int \cos x\,dx = \sin x + C$ and $\int \frac{1}{x}\,dx = \ln x +C$ are elementary, but $\int \frac{\cos x}{x}\,dx$ is not.
A: Arguably integration by parts is the anti-derivative form of the product rule; in fact it is derived from the differentiation (i.e. the usual) product rule.
A: HINT $\ $  If there were a product rule $\rm\:\int f\:g\ =\ p(f,g,\int f,\int g)\:$ for some polynomial $\rm\:p\:$ then for $\rm\:f = x,\ g = 1/x^2\:$ this would imply that $\rm\: log(x)\:$ is a rational function.
A: 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.
