# How do I enter (the arguments of) this simple Integral?

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 $$\int \frac{1}{(x y)^3} \, d(x y)=-\frac{1}{2 x^2 y^2}$$In[1249]:= Integrate[1/(x y)^3, x, y] (* wrong syntax for what i wish, d(xy) *)Out[1249]= 1/(4 x^2 y^2) (* wrong per say - not the solution sought *)(the above is a double integral, the of inverse of two partial derivatives). I'm looking for answer:-1/(2 x^2 y^2) In[1240]:= Integrate[x y, {x, y}] During evaluation of In[1240]:= Integrate::ilim: Invalid integration variable or limit(s) in {x,y}. Out[1240]= Integrate[x y, {x, y}] In[1242]:= Integrate[x y, x y] During evaluation of In[1242]:= Integrate::ilim: Invalid integration variable or limit(s) in x y.  my well respected book reads: d(xy)+x^3y^2 dy=0 (6) Since the differential xy is present in equation (6), any factor that is a function of the product xy will not disturb the integrability of that term. But the other term contains the differential dy, and hence should contain a function of y alone. Therefore, let us divide by (xy)^3 and write d(xy)/(xy)^3 + dy/y = 0 the equation above is integrable as it stands. a family of solutions is defined by: -1/(2x^2y^2) + ln|y| = - ln|c|  how did d(xy) get in the equation?on the page before a list of "differentials that occur frequently" is given (useful for substitution), one of is: d(xy) = x dy + y dx before the snip further above, that substitution had already been made, which is how d(xy) entered into the equationmm knows some substitutions like transcendental ones. i'm not sure it knows/uses/needs differential ones