# Euler substitution

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vte :
Euler substitution is a method for evaluating integrals of the form
$\int R(x,{\sqrt {ax^{2}+bx+c}})\,dx,$
where $R$ is a rational function of $x$ and √(ax$^2$+bx+c){\textstyle {\sqrt {ax^{2}+bx+c}}}. In such cases, the integrand can be changed to a rational function by using the substitutions of Euler.

## Euler's first substitution

The first substitution of Euler is used when $a>0$. We substitute ${\sqrt {ax^{2}+bx+c}}=\pm x{\sqrt {a}}+t$ and solve the resulting expression for $x$. We have that $x={\frac {c-t^{2}}{\pm 2t{\sqrt {a}}-b}}$ and that the $dx$ term is expressible rationally in $t$.
In this substitution, either the positive sign or the negative sign can be chosen.

## Euler's second substitution

If $c>0$, we take ${\sqrt {ax^{2}+bx+c}}=xt\pm {\sqrt {c}}.$ We solve for $x$ similarly as above and find $x={\frac {\pm 2t{\sqrt {c}}-b}{a-t^{2}}}.$
Again, either the positive or the negative sign can be chosen.

## Euler's third substitution

If the polynomial $ax^{2}+bx+c$ has real roots $\alpha$ and $\beta$, we may choose √(ax$^2$+bx+c)=√(a(x−α)(x−β))=(x−α)t{\textstyle {\sqrt {ax^{2}+bx+c}}={\sqrt {a(x-\alpha )(x-\beta )}}=(x-\alpha )t}. This yields $x={\frac {a\beta -\alpha t^{2}}{a-t^{2}}},$ and as in the preceding cases, we can express the entire integrand rationally in $t$.

## Worked examples
### Examples for Euler's first substitution
#### One

In the integral $\int \!{\frac {\ dx}{\sqrt {x^{2}+c}}}$ we can use the first substitution and set √(x$^2$+c)=−x+t{\textstyle {\sqrt {x^{2}+c}}=-x+t}, thus $x={\frac {t^{2}-c}{2t}}\quad \quad \ dx={\frac {t^{2}+c}{2t^{2}}}\,\ dt$ ${\sqrt {x^{2}+c}}=-{\frac {t^{2}-c}{2t}}+t={\frac {t^{2}+c}{2t}}$ Accordingly, we obtain: $\int {\frac {dx}{\sqrt {x^{2}+c}}}=\int {\frac {\frac {t^{2}+c}{2t^{2}}}{\frac {t^{2}+c}{2t}}}\,\ dt=\int {\frac {dt}{t}}=\ln |t|+C=\ln \left|x+{\sqrt {x^{2}+c}}\right|+C$
The cases $c=\pm 1$ give the formulas ${\begin{aligned}\int {\frac {\ dx}{\sqrt {x^{2}+1}}}&=\operatorname {arsinh} (x)+C\\[6pt]\int {\frac {\ dx}{\sqrt {x^{2}-1}}}&=\operatorname {arcosh} (x)+C\qquad (x>1)\end{aligned}}$

#### Two

For finding the value of $\int {\frac {1}{x{\sqrt {x^{2}+4x-4}}}}dx,$ we find $t$ using the first substitution of Euler, √(x$^2$+4x−4)=√1x+t=x+t{\textstyle {\sqrt {x^{2}+4x-4}}={\sqrt {1}}x+t=x+t}. Squaring both sides of the equation gives us $x^{2}+4x-4=x^{2}+2xt+t^{2}$, from which the $x^{2}$ terms will cancel out. Solving for $x$ yields $x={\frac {t^{2}+4}{4-2t}}.$
From there, we find that the differentials $dx$ and $dt$ are related by $dx={\frac {-2t^{2}+8t+8}{(4-2t)^{2}}}dt.$
Hence, ${\begin{aligned}\int {\frac {dx}{x{\sqrt {x^{2}+4x-4}}}}&=\int {\frac {\frac {-2t^{2}+8t+8}{(4-2t)^{2}}}{\left({\frac {t^{2}+4}{4-2t}}\right)\left({\frac {-t^{2}+4t+4}{4-2t}}\right)}}dt&&t={\sqrt {x^{2}+4x-4}}-x\\[6pt]&=2\int {\frac {dt}{t^{2}+4}}=\tan ^{-1}\left({\frac {t}{2}}\right)+C\\[6pt]&=\tan ^{-1}\left({\frac {{\sqrt {x^{2}+4x-4}}-x}{2}}\right)+C\end{aligned}}$

### Examples for Euler's second substitution

In the integral $\int \!{\frac {dx}{x{\sqrt {-x^{2}+x+2}}}},$ we can use the second substitution and set ${\sqrt {-x^{2}+x+2}}=xt+{\sqrt {2}}$. Thus $x={\frac {1-2{\sqrt {2}}t}{t^{2}+1}}\qquad dx={\frac {2{\sqrt {2}}t^{2}-2t-2{\sqrt {2}}}{(t^{2}+1)^{2}}}dt,$ and ${\sqrt {-x^{2}+x+2}}={\frac {1-2{\sqrt {2}}t}{t^{2}+1}}t+{\sqrt {2}}={\frac {-{\sqrt {2}}t^{2}+t+{\sqrt {2}}}{t^{2}+1}}$
Accordingly, we obtain: ${\begin{aligned}\int {\frac {dx}{x{\sqrt {-x^{2}+x+2}}}}&=\int {\frac {\frac {2{\sqrt {2}}t^{2}-2t-2{\sqrt {2}}}{(t^{2}+1)^{2}}}{{\frac {1-2{\sqrt {2}}t}{t^{2}+1}}{\frac {-{\sqrt {2}}t^{2}+t+{\sqrt {2}}}{t^{2}+1}}}}dt\\[6pt]&=\int \!{\frac {-2}{-2{\sqrt {2}}t+1}}dt={\frac {1}{\sqrt {2}}}\int {\frac {-2{\sqrt {2}}}{-2{\sqrt {2}}t+1}}dt\\[6pt]&={\frac {1}{\sqrt {2}}}\ln \left|2{\sqrt {2}}t-1\right|+C\\[4pt]&={\frac {\sqrt {2}}{2}}\ln \left|2{\sqrt {2}}{\frac {{\sqrt {-x^{2}+x+2}}-{\sqrt {2}}}{x}}-1\right|+C\end{aligned}}$

### Examples for Euler's third substitution

To evaluate $\int \!{\frac {x^{2}}{\sqrt {-x^{2}+3x-2}}}\ dx,$ we can use the third substitution and set √(−(x−2)(x−1))=(x−2)t{\textstyle {\sqrt {-(x-2)(x-1)}}=(x-2)t}. Thus $x={\frac {-2t^{2}-1}{-t^{2}-1}}\qquad \ dx={\frac {2t}{(-t^{2}-1)^{2}}}\,\ dt,$ and ${\sqrt {-x^{2}+3x-2}}=(x-2)t={\frac {t}{-t^{2}-1.}}$
Next, $\int {\frac {x^{2}}{\sqrt {-x^{2}+3x-2}}}\ dx=\int {\frac {\left({\frac {-2t^{2}-1}{-t^{2}-1}}\right)^{2}{\frac {2t}{(-t^{2}-1)^{2}}}}{\frac {t}{-t^{2}-1}}}\ dt=\int {\frac {2(-2t^{2}-1)^{2}}{(-t^{2}-1)^{3}}}\ dt.$ As we can see this is a rational function which can be solved using partial fractions.

## Generalizations

The substitutions of Euler can be generalized by allowing the use of imaginary numbers. For example, in the integral ∫ dx/(√(−x$^2$+c)) {\textstyle \int {\frac {dx}{\sqrt {-x^{2}+c}}}}, the substitution √(−x$^2$+c)=±ix+t{\textstyle {\sqrt {-x^{2}+c}}=\pm ix+t} can be used. Extensions to the complex numbers allows us to use every type of Euler substitution regardless of the coefficients on the quadratic.
The substitutions of Euler can be generalized to a larger class of functions. Consider integrals of the form $\int R_{1}\left(x,{\sqrt {ax^{2}+bx+c}}\right)\,\log \left(R_{2}\left(x,{\sqrt {ax^{2}+bx+c}}\right)\right)\,dx,$ where $R_{1}$ and $R_{2}$ are rational functions of $x$ and √(ax$^2$+bx+c){\textstyle {\sqrt {ax^{2}+bx+c}}}. This integral can be transformed by the substitution √(ax$^2$+bx+c)=√a+xt{\textstyle {\sqrt {ax^{2}+bx+c}}={\sqrt {a}}+xt} into another integral $\int {\tilde {R}}_{1}(t)\log {\big{\tilde {R}}_{2}(t){\big )}\,dt,$ where ${\tilde {R}}_{1}(t)$ and ${\tilde {R}}_{2}(t)$ are now simply rational functions of $t$. In principle, factorization and partial fraction decomposition can be employed to break the integral down into simple terms, which can be integrated analytically through use of the dilogarithm function.