In the integration task $$\int\!R(x,\,\sqrt{ax^2+bx+c})\,dx,$$ 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

  (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

  (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

  (3)

  
  Now $$ax^2+bx+c = a(x-\alpha)(x-\beta) = (x-\alpha)^2t^2,$$ 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 $$x = \frac{t^2-c}{2t},\;\; dx = \frac{t^2+c}{2t^2}\,dt,\;\; \sqrt{x^2+c} = -\frac{t^2-c}{2t}+t = \frac{t^2+c}{2t}.$$ Accordingly we obtain $$\int\!\frac{dx}{\sqrt{x^2+c}} = \int\!\frac{\frac{t^2+c}{2t^2}dt}{\frac{t^2+c}{2t}} = \int\!\frac{dt}{t} = \ln|t|+C = \ln|x+\sqrt{x^2+c}|+C.$$ Especially the cases $c = \pm1$ give the formulas $$\int\!\frac{dx}{\sqrt{x^2+1}} = \arsinh{x}+C, \quad \int\!\frac{dx}{\sqrt{x^2-1}} = \arcosh{x}+C \;\; (x > 1).$$

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 $c^2-x^2 = x^2t^2-2cxt+c^2$ , which implies $$x = \frac{2ct}{t^2+1},\;\; dx = \frac{2c(1-t^2)dt}{(1+t^2)^2},\;\; \sqrt{c^2-x^2} = \frac{2ct^2}{t^2+1}-c = \frac{c(t^2-1)}{t^2+1}.$$ We then obtain $$\int\!\frac{\sqrt{c^2-x^2}}{x}\,dx = -c\int\!\frac{(1-t^2)^2}{t(1+t^2)^2}\,dt = c\int\!\left(\frac{4t}{(1+t^2)^2}-\frac{1}{t}\right)dt = -\frac{2c}{1+t^2}-c\ln|t|+C_1.$$ 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 $$\int\!\frac{\sqrt{c^2-x^2}}{x}\,dx = -\frac{x^2}{c+\sqrt{c^2-x^2}}-c\ln\frac{c+\sqrt{c^2-x^2}}{x}+C_1 = -c+\sqrt{c^2-x^2}-c\ln\frac{c+\sqrt{c^2-x^2}}{x}+C_1,$$ i.e. $$\int\!\frac{\sqrt{c^2-x^2}}{x}\,dx = \sqrt{c^2-x^2}-c\ln\frac{c+\sqrt{c^2-x^2}}{x}+C.$$

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 $$x = \frac{1+4t^2}{1-t^2},\;\; dx = \frac{10t}{(1-t^2)^2}\,dt,\;\; \sqrt{x^2+3x-4} = \left(\frac{1+4t^^2}{1-t^2}+4\right)t = \frac{5t}{1-t^2}.$$ And we obtain $$\int\!\frac{dx}{\sqrt{x^2+3x-4}} = \int\!\frac{10t(1-t^2)}{(1-t^2)^2\cdot5t}\,dt = \int\!\frac{2}{1-t^2}\,dt = \ln\left|\frac{1+t}{1-t}\right|+C = \ln\left|\frac{1+\sqrt{\frac{x-1}{x+4}}}{1-\sqrt{\frac{x-1}{x+4}}}\right|+C$$ $$= \ln\left|\frac{\sqrt{x+4}+\sqrt{x-1}}{\sqrt{x+4}-\sqrt{x-1}}\right|+C.$$

Bibliography

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N. PISKUNOV: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele. Viies, täiendatud trükk. Kirjastus ``Valgus'', Tallinn (1965).