Euler’s substitutions for integration

In the integration task

$$\int R(x,\sqrt{a{x}^2+bx+c})𝑑x,$$

where the integrand is a rational function of $x$ and $\sqrt{a{x}^2+bx+c}$, 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

  $\sqrt{a{x}^2+bx+c}=\pm x\sqrt{a}+t.$(1)

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

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

  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

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

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

  $\sqrt{a{x}^2+bx+c}=xt\pm \sqrt{c}.$(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 $a{x}^2+bx+c$ has the real zeros $\alpha$ and $\beta$, we may chose

  $\sqrt{a{x}^2+bx+c}=(x-\alpha )t.$(3)

  Now

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

  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 $\sqrt{a{x}^2+bx+c}$ 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  $c=\pm 1$  give the formulas

$$\int \frac{dx}{\sqrt{x^2+1}}=\mathrm{arsinh}x+C,\int \frac{dx}{\sqrt{x^2-1}}=\mathrm{arcosh}x+C(x>1).$$

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  $c^2-x^2=x^2{t}^2-2cxt+c^2$, which implies

$$x=\frac{2ct}{t^2+1},dx=\frac{2c(1-t^2)dt}{{(1+t^2)}^2},\sqrt{c^2-x^2}=\frac{2c{t}^2}{t^2+1}-c=\frac{c(t^2-1)}{t^2+1}.$$

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

$$x=\frac{1+4t^2}{1-t^2},dx=\frac{10t}{{(1-t^2)}^2}dt,\sqrt{x^2+3x-4}=\left(\frac{1+4tr}{1-t^2}+4\right)t=\frac{5t}{1-t^2}.$$

And we obtain

$$\int \frac{dx}{\sqrt{x^2+3x-4}}=\int \frac{10t(1-t^2)}{{(1-t^2)}^2\cdot 5t}𝑑t=\int \frac{2}{1-t^2}𝑑t=\ln \left|\frac{1+t}{1-t}\right|+C=\ln \left|\frac{1+\sqrt{\frac{x-1}{x+4}}}{1-\sqrt{\frac{x-1}{x+4}}}\right|+C$$

$$=\ln \left|\frac{\sqrt{x+4}+\sqrt{x-1}}{\sqrt{x+4}-\sqrt{x-1}}\right|+C.$$