Details

Consider the curve (1) and a point on it. The straight line (2) through intersects the curve in another point . Eliminating from (1) and (2) gives

(3).

From that, and since

,

(3) becomes

,

which simplifies to

.

So is a rational function of , is a rational function of , and because of (2), is a rational function of .

So the relation

defines the substitution that rationalizes the integral.

Suppose that the trinomial has a real root . Then we get Euler's second substitution taking ,

.

If , then the curve intersects the axis at , which must be the point . This is Euler's third substitution

.

In the case of Euler's first substitution, the point is at infinity, , so the curve is a hyperbola. An asymptote is . We are looking for the intersection of the curve by straight lines that are parallel to the asymptote. The intersection of such a line gives a point , which is rational in terms of . This gives Euler's first substitution

.

Reference

[1] G. M. Fihtenholjc, Lectures in Differential and Integral Calculus (in Russian), Vol. 2, Moscow: Nauka, 1966 pp. 56–60.