How could I proceed in order to solve this double integral?

$$\iint_D \frac{3+ e^{yx} -yxe^{yx}}{(3+e^{yx})^2}dxdy $$ where D is the region whose boundary is the square with vertices (0,0), (1,0), (1,1), (0,1).

I have though about change of variables, but I can only think of having u = xy, no clue for v.

All clues are appreciated.

  • $\begingroup$ It sort of looks like that's the derivative of a quotient. Following through with that, we could see that the integrand is the derivative of the function $\frac{xy}{3+e^{xy}}$ with respect to $xy$. $\endgroup$ – Harry May 12 '17 at 7:48
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    $\begingroup$ The boundary of a square is a closed curve (one-dimensional), and not a two-dimensional region on which one can take an area integral. So is D perhaps the square region (not its boundary) ? $\endgroup$ – Manuel Guillen May 12 '17 at 8:22
  • $\begingroup$ Thanks Manuel, I will update that. I must have copied this wrong. I finally came to the conclusion that it was a quotient integral, but I was a little skeptical about not finding any other way to solve the integral. Thanks to both of you! $\endgroup$ – Bee May 12 '17 at 16:37

There is no need to set $u=xy$, because we have:

\begin{align} \int_0^1\int_0^1 \frac{3+e^{xy}-yxe^{yx}} {(3+e^{yx})^2}dxdy =\int_0^1 \bigg[\frac{x} {3+e^{xy}} \bigg]^1_0dy=\int_0^1\frac{1}{3+e^y}dy\end{align} The last integral can be solved easily if you rewrite it as: \begin{align} \int_0^1\frac{1}{3+e^y}dy = \int_0^1\frac{e^{-y} }{3e^{-y}+1}dy\end{align} I hope this helped.


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