Computing a Differential Form Apologies in advance, I don't know TeX, so this might look a bit gross...
I'm given a 1-form $A=f_1dx_1+...+f_ndx_n$, infinitely differentiable and closed on $R^n$.
I want to show that $dg=A$ for 
$$ g= \int_{0}^{x_1}f_1(t,x_2,x_3,...,x_n)dt+\int_{0}^{x_2}f_2(0,t,x_3,...,x_n)dt+...+\int_{0}^{x_n}f_n(0,...,0,t)dt. $$
I thought computing $dg$ was straightforward, but the hint makes me think I'm abusing some elementary calculus: "apply the fundamental theorem of calculus, differentiate under the integral sign, and don't forget to use that $dA=0$."
Any and all help is much appreciated.  Again, so sorry for the formatting...
Edit: So I know that the partial of $f_i$ with respect to $x_j$ equals the partial with the indices switched, and applying FTC certainly now gets rid of all the integrals, but I still can't quite figure out how all the extraneous stuff cancels nicely...
 A: I calculate $dg = f_1dx_1+\cdots f_ndx_n$, but the arguments of $f_1,f_2, \dots f_n$ would seem to require some thinking:
Let me explain how this happens,
\begin{align}
g(x_1,x_2, \dots , x_n) &= \int_{0}^{x_1}f_1(t,x_2, \dots ,x_n)dt 
 + \int_{0}^{x_2}f_1(0,t,x_3, \dots ,x_n)dt + \\
& \qquad  \cdots + \int_{0}^{x_n}f_1(0,0, \dots ,t)dt 
\end{align}
Differentiating, the $j$-term is nontrivial by the FTC
$$ \frac{\partial g}{\partial x_j} = \frac{\partial}{\partial x_j} \int_{0}^{x_j}f_j(0, \dots ,t,x_{j+1}, \dots , x_n)dt 
=f_j(0, \dots, x_j, x_{j+1}, \dots , x_n)+ \star
$$
However, where $\star$ is from the remaining terms $f_k$ with $k<j$ also have an $x_j$-dependence in integrand
$$ \star = \sum_{i=1}^{j-1}\int_{0}^{x_i}\frac{\partial f_i}{\partial x_j}(0,\dots, t,x_{i+1}, \dots x_j, \dots, x_n)dt$$
I wonder, did you notice these terms? These require the other FTC.
added details: let me elaborate on the calculation:
$$ \frac{\partial g}{\partial x_1} = f_1(x_1,\dots , x_n) $$
\begin{align}
 \frac{\partial g}{\partial x_2} &= f_2(0,x_2,\dots , x_n) +
\frac{\partial }{\partial x_2} \int_{0}^{x_1}f_1(t,x_2, \dots x_n)dt \\
&= f_2(0,x_2,\dots , x_n) +
 \int_{0}^{x_1}\frac{\partial f_1 }{\partial x_2}(t,x_2, \dots x_n)dt \\
&= f_2(0,x_2,\dots , x_n) +
 \int_{0}^{x_1}\frac{\partial f_2 }{\partial x_1}(t,x_2, \dots x_n)dt \\
&= f_2(0,x_2,\dots , x_n) +
  f_2(x_1,x_2, \dots x_n)-f_2(0,x_2, \dots x_n) \\
&=f_2(x_1,x_2, \dots x_n).
\end{align}
This starts to be a pain to write explicitly. However, if this doesn't make sense then it is probably for neglect of those evaluations.
\begin{align}
 \frac{\partial g}{\partial x_3} &= f_3(0,0,x_3,\dots , x_n) +
\frac{\partial }{\partial x_3} \int_{0}^{x_1}f_1 \, dt +
\frac{\partial }{\partial x_3} \int_{0}^{x_2}f_2 \, dt \\
&= f_3(x_1,\dots ,x_n)
\end{align}
In the last step I omitted a pair of $f_3(0,0,x_3, \dots, x_n)$ which cancel and a pair of $f_3(0,x_2,x_3, \dots, x_n)$ which cancel. This pattern continues to higher terms. For example, in the computation of $\frac{\partial g}{\partial x_4}$ there are 7 nontrivial terms and only $f_4(x_1,\dots ,x_n)$ remains.
In short, I think my previous version was correct. There is always one term arising from the varying bound and an even number of terms coming from bringing in the derivative (assumes uniform continuity of $f_j$) into the integral. As the computation unfolds everything cancels modulo the term $f_j(x_1, \dots x_n)$.
My future students do not thank you for this homework :)
