Integral as weighted sum of derivatives. Is this a new result? $$\int f(x) \, dx = \sum_{n=1}^\infty (-1)^{n+1}*\frac{x^n}{n!}\frac{d^nf(x)}{dx^n}$$
I derived this equation from the repeated application of the chain rule.
$$\int f(x) \, dx = \int 1*f(x) \, dx$$
$$=x*f(x)-\int x*f'(x) \, dx$$
$$=x*f(x)-(\frac{x^2*f'(x)}{2}-\int\frac{x^2}{2} f''(x) \, dx)$$
at this point I just assumed the pattern continues and wrote the equation.
I posted this before (with an error) trying to find a source with a similar result. It is awfully similar to McClaurin series (as somebody pointed out on my previous post), however it IS different. I was unable to find this equation from Google. I would appreciate if somebody could link a source with this result!
 A: To start with, the issue with your formula is that it treats the antiderivative symbol as if it was an operator acting on $f,$ but this is problematic, because the antiderivative of a function $f$ is not unique. Let us change the starting point to an integral, $$\int_a^tf(x)\,\mathrm{d}x.$$ Assume $f\in{C^1(I)},$ that is, assume $f$ is continuously differentiable once on the open interval $I$ containing $a.$ Then it follows that $$\int_a^tf(x)\,\mathrm{d}x=tf(t)-af(a)-\int_a^txD[f](x)\,\mathrm{d}x.$$ I am using $D$ to denote the derivative operator, as I find this to be more convenient than the prime notation. If $f\in{C^2(I)},$ then $$\int_a^tf(x)\,\mathrm{d}x=tf(t)-af(a)-\left[\frac{t^2}2D[f](t)-\frac{a^2}2D[f](a)\right]+\int_a^txD[f](x)\,\mathrm{d}x=\sum_{m=0}^{2-1}(-1)^m\frac{t^{m+1}}{m!}D^m[f](t)-\sum_{m=0}^{2-1}(-1)^m\frac{a^{m+1}}{m!}D^m[f](a)+(-1)^2\int_a^t\frac{x^2}{2!}D^2[f](x)\,\mathrm{d}x.$$ You can continue this pattern, and for $f\in{C^n(I)},$ you can use induction to rigorously prove that $$\int_a^tf(x)\,\mathrm{d}x=\sum_{m=0}^{n-1}(-1)^m\frac{t^{m+1}}{m!}D^m[f](t)-\sum_{m=0}^{n-1}(-1)^m\frac{a^{m+1}}{m!}D^m[f](a)+(-1)^n\int_a^t\frac{x^n}{n!}D^n[f](x)\,\mathrm{d}x.$$ Instead, if you assume that $f\in{C^{\infty}(I)},$ then you can consider letting $n\to\infty,$ hence $$\int_a^tf(x)\,\mathrm{d}x=\sum_{m=0}^{\infty}(-1)^m\frac{t^{m+1}}{m!}D^m[f](t)-\sum_{m=0}^{\infty}(-1)^m\frac{a^{m+1}}{m!}D^m[f](a)+\lim_{n\to\infty}(-1)^n\int_a^t\frac{x^n}{n!}D^n[f](x)\,\mathrm{d}x.$$ There may be some $f\in{C^{\infty}(I)}$ with special convergence properties, such that $$\lim_{n\to\infty}(-1)^n\int_a^t\frac{x^n}{n!}D^n[f](x)\,\mathrm{d}x=0.$$ In that case, $$\int_a^tf(x)\,\mathrm{d}x=\sum_{m=0}^{\infty}(-1)^m\frac{t^{m+1}}{m!}D^m[f](t)-\sum_{m=0}^{\infty}(-1)^m\frac{a^{m+1}}{m!}D^m[f](a),$$ which I think is what you are getting at. However, notice all the assumptions that had to be made to get here, and in particular, that last convergence condition is very strong, so not many functions would satisfy it.
That being said, there is further analysis to be made here. Notice that, per Taylor's theorem, $$f(x+a)=\sum_{m=0}^{\infty}\frac{a^m}{m!}D^m[f](x)$$ for functions that are sufficiently smooth. So, in fact, $$\sum_{m=0}^{\infty}(-1)^m\frac{a^{m+1}}{m!}D^m[f](a)=af[a+(-a)]=af(0),$$ at least in the case that both $a$ and $-a$ are in $I.$ Also, the series expansion must converge, which is not necessarily true. Regardless, this means that, provided all the assumptions made above, $$\int_a^tf(x)\,\mathrm{d}x=(t-a)f(0).$$ But notice that $$(t-a)f(0)=\int_a^tx\,\mathrm{d}xf(0)=\int_a^txf(0)\,\mathrm{d}x.$$ This is a very strong property, and again, very few functions satisfy it.
That is the real problem with your formula. It naively makes many strong assumptions stacked on top of each other. So the formula is true for only a "small" class of functions.
