In the integration task

where the integrand is a rational function of $x$ and $\sqrt{ax^{2}+bx+c}$, the integrand can be changed to a rational function of a new variable $t$ by using the following substitutions of Euler.

• The first substitution of Euler.  If  $a>0$,  we may write

  $\displaystyle\sqrt{ax^{2}+bx+c}\;=\;\pm x\sqrt{a}+t.$

  (1)

  When we take $\sqrt{a}$ with the minus sign, then

  $$ax^{2}+bx+c\;=\;ax^{2}-2xt\sqrt{a}+t^{2},$$

  from which we get the expression

  $$x\;=\;\frac{t^{2}-c}{b+2t\sqrt{a}};$$

  thus also $dx$ is expressible rationally via $t$. We have

  $$\sqrt{ax^{2}+bx+c}\;=\;-x\sqrt{a}+t\;=\;\frac{c-t^{2}}{b+2t\sqrt{a}}\sqrt{a}+t.$$

• The second substitution of Euler.  If  $c>0$,  we take

  $\displaystyle\sqrt{ax^{2}+bx+c}\;=\;xt\pm\sqrt{c}.$

  (2)

  With the minus sign we obtain, similarly as above,

  $$x\;=\;\frac{2t\sqrt{c}+b}{t^{2}-a}.$$

• The third substitution of Euler.  If the polynomial $ax^{2}\!+\!bx\!+\!c$ has the real zeros $\alpha$ and $\beta$, we may chose

  $\displaystyle\sqrt{ax^{2}\!+\!bx\!+\!c}\;=\;(x\!-\!\alpha)t.$

  (3)

  Now

  $$ax^{2}\!+\!bx\!+\!c\;=\;a(x\!-\!\alpha)(x\!-\!\beta)\;=\;(x\!-\!\alpha)^{2}t^{2},$$

  whence  $a(x\!-\!\beta)=(x\!-\!\alpha)t^{2}$. This gives the expression

  $$x\;=\;\frac{a\beta\!-\!\alpha t^{2}}{a\!-\!t^{2}}.$$

  As in the preceding cases, we can express $dx$ and $\sqrt{ax^{2}\!+\!bx\!+\!c}$ rationally via $t$.

Examples.

1. In the integral $\displaystyle\int\!\frac{dx}{\sqrt{x^{2}+c}}$ we can use the first substitution:  $\sqrt{x^{2}+c}=-x+t$;  then  $x^{2}+c=x^{2}-2xt+t^{2}$  and thus

Accordingly we obtain

Especially the cases  $c=\pm 1$  give the formulas

2. The integral $\displaystyle\int\!\frac{\sqrt{c^{2}-x^{2}}}{x}\,dx$ is needed in deriving the equation of the tractrix. We use for integrating the second substitution  $\sqrt{c^{2}-x^{2}}=xt-c$;  then  $c^{2}-x^{2}=x^{2}t^{2}-2cxt+c^{2}$, which implies

We then obtain

The equation tying $x$ and $t$ gives  $\frac{2c}{1+t^{2}}=\frac{x}{t}$  and  $t=\frac{c+\sqrt{c^{2}-x^{2}}}{x}$,  whence

i.e.

3. In the integral $\displaystyle\int\!\frac{dx}{\sqrt{x^{2}+3x-4}}$, the radicand is $(x+4)(x-1)$. Using the third substitution of Euler, we take  $\sqrt{x^{2}+4x-3}=(x+4)t$. This simplifies to  $x-1=(x+4)t^{2}$. Then we get

And we obtain