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Here's an integral which I encountered that uses some unfamiliar notation for me: $$\int-\frac{d(x/y)}{\sqrt{1+(x/y)^2}}$$ What does this mean? I don't have much of an idea.

Edit: This problem is from a book on differential equations ($y$ is a function), and the author writes:

$$-\frac{d(x/y)}{\sqrt{1+(x/y)^2}}=\frac{dx}{x}$$ Integration of this now gives $$-\log\left|\frac xy +\sqrt{1+(x/y)^2}\right|=\log|x|+\log|c|$$

How do you get to this step?

Tdonut
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2 Answers2

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Multiple choice:

  • $y$ is a constant, i.e. something that doesn't change as $x$ changes, and you're antidifferentiating a function of $x$. Then this becomes $$ -\int\frac{du}{\sqrt{1+u^2}}. $$
  • It's not altogether impossible that $x$ is a constant and you're antidifferentiating a function of $y$, so that $\displaystyle d\left(\frac x y \right) = \frac{-x}{y^2}\,dy $
  • $x$ and $y$ are related to each other in some way specified elsewhere than in what is posted in the question above. If, for example, both are functions of $t$, then we could have $$d\left(\frac x y \right) = \frac{y\,dx-x\,dy}{y^2} = \frac{yx'-xy'}{y^2}\,dt,$$ where $x'$ means $dx/dt$ and $y'$ means $dy/dt$.
  • Maybe something else?

(The third bullet point above actually includes both of the first two.)

I wouldn't use such notation without at least having some preceding context that would make it clear which choice above is right, and even then maybe I wouldn't be too comfortable with it.

Michael Hardy
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  • One question: wouldn't all these approaches in the end lead to the same result? – Fabian May 21 '15 at 19:30
  • @Fabian : If $(x,y)$ follows some path through the plane as $t$ varies, then the third bullet point makes sense but the first and second don't unless the path is a line parallel to one of the coordinate axes, and then only one of the first two would be right. ${}\qquad{}$ – Michael Hardy May 21 '15 at 19:38
  • What I am trying to say is that the line-integral does not depend on the path but only on the end-points. In fact it only depends on the ratio $y/x$ of the starting point and the end point. – Fabian May 21 '15 at 20:36
  • @Fabian : But neither of the first two approaches would give the right result unless either $x$ or $y$ is constant. – Michael Hardy May 21 '15 at 20:39
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The notation is a bit unclear to me. But let's say that you want to find the following integral. $$ \int -\frac{1}{\sqrt{1 + (x/y)^2}}\; dx. $$ Here, from the notation, we assume that $y$ is a constant. The standard approach is integration by substitution. So we let $u = x/y$ Then $du = (1/y) dy$. So we get the integral $$ \int -\frac{1}{\sqrt{1 + (u)^2}}\; y\;du. $$ This we can find. (something with an inverse trig. function).

Another notation for the above is the notation that you ask about. It is, however, a bit more unclear in my opinion.

Anyway, the point is that this notation is an "alternative" to integration by substitution.

Thomas
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