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

Posted 1 month ago
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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 equation

mm knows some substitutions like transcendental ones. i'm not sure it knows/uses/needs differential ones

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