Euler substitution

In the integral ${\displaystyle \int \!{\frac {\ dx}{\sqrt {x^{2}+c}}}}$ we can use the first substitution and set ${\displaystyle {\sqrt {x^{2}+c}}=-x+t}$, thus

${\displaystyle x={\frac {t^{2}-c}{2t}}\quad \quad \ dx={\frac {t^{2}+c}{2t^{2}}}\,\ dt}$

${\displaystyle {\sqrt {x^{2}+c}}=-{\frac {t^{2}-c}{2t}}+t={\frac {t^{2}+c}{2t}}}$

Accordingly, we obtain:

${\displaystyle \int {\frac {\ dx}{\sqrt {x^{2}+c}}}=\int {\frac {\frac {t^{2}+c}{2t^{2}}}{\frac {t^{2}+c}{2t}}}\,\ dt=\int \!{\frac {\ dt}{t}}=\ln |t|+C=\ln |x+{\sqrt {x^{2}+c}}|+C}$

The cases ${\displaystyle c=\pm 1}$ give the formulas

${\displaystyle {\begin{aligned}\int {\frac {\ dx}{\sqrt {x^{2}+1}}}&={\mbox{arsinh}}(x)+C\\[6pt]\int {\frac {\ dx}{\sqrt {x^{2}-1}}}&={\mbox{arcosh}}(x)+C\qquad (x>1)\end{aligned}}}$

For finding the value of

${\displaystyle \int {\frac {1}{x{\sqrt {x^{2}+4x-4}}}}dx,}$

we find ${\displaystyle t}$ using the first substitution of Euler, ${\displaystyle {\sqrt {x^{2}+4x-4}}={\sqrt {1}}x+t=x+t}$. Squaring both sides of the equation gives us ${\displaystyle x^{2}+4x-4=x^{2}+2xt+t^{2}}$, from which the ${\displaystyle x^{2}}$ terms will cancel out. Solving for ${\displaystyle x}$ yields

${\displaystyle x={\frac {t^{2}+4}{4-2t}}.}$

From there, we find that the differentials ${\displaystyle dx}$ and ${\displaystyle dt}$ are related by

${\displaystyle dx={\frac {-2t^{2}+8t+8}{(4-2t)^{2}}}dt.}$

Hence,

${\displaystyle {\begin{aligned}\int {\frac {dx}{x{\sqrt {x^{2}+4x-4}}}}&=\int {\frac {\frac {-2t^{2}+8t+8}{(4-2t)^{2}}}{({\frac {t^{2}+4}{4-2t}})({\frac {-t^{2}+4t+4}{4-2t}})}}dt\\[6pt]&=2\int {\frac {dt}{t^{2}+4}}=\tan ^{-1}\left({\frac {t}{2}}\right)+C&&t={\sqrt {x^{2}+4x-4}}-x\\[6pt]&=\tan ^{-1}\left({\frac {{\sqrt {x^{2}+4x-4}}-x}{2}}\right)+C\end{aligned}}}$

In the integral

${\displaystyle \int \!{\frac {dx}{x{\sqrt {-x^{2}+x+2}}}},}$

we can use the second substitution and set ${\displaystyle {\sqrt {-x^{2}+x+2}}=xt+{\sqrt {2}}}$. Thus

${\displaystyle x={\frac {1-2{\sqrt {2}}t}{t^{2}+1}}\qquad dx={\frac {2{\sqrt {2}}t^{2}-2t-2{\sqrt {2}}}{(t^{2}+1)^{2}}}dt,}$

and

${\displaystyle {\sqrt {-x^{2}+x+2}}={\frac {1-2{\sqrt {2t}}}{t^{2}+1}}t+{\sqrt {2}}={\frac {-{\sqrt {2}}t^{2}+t+{\sqrt {2}}}{t^{2}+1}}}$

Accordingly, we obtain:

${\displaystyle {\begin{aligned}\int {\frac {dx}{x{\sqrt {-x^{2}+x+2}}}}&=\int {\frac {\frac {2{\sqrt {2}}t^{2}-2t-2{\sqrt {2}}}{(t^{2}+1)^{2}}}{{\frac {1-2{\sqrt {2}}t}{t^{2}+1}}{\frac {-{\sqrt {2}}t^{2}+t+{\sqrt {2}}}{t^{2}+1}}}}dt\\[6pt]&=\int \!{\frac {-2}{-2{\sqrt {2}}t+1}}dt={\frac {1}{\sqrt {2}}}\int {\frac {-2{\sqrt {2}}}{-2{\sqrt {2}}t+1}}dt\\[6pt]&={\frac {1}{\sqrt {2}}}\ln {\Biggl |}2{\sqrt {2}}t-1{\Biggl |}+C={\frac {\sqrt {2}}{2}}\ln {\Biggl |}2{\sqrt {2}}{\frac {{\sqrt {-x^{2}+x+2}}-{\sqrt {2}}}{x}}-1{\Biggl |}+C\end{aligned}}}$

To evaluate

${\displaystyle \int \!{\frac {x^{2}}{\sqrt {-x^{2}+3x-2}}}\ dx,}$

we can use the third substitution and set ${\displaystyle {\sqrt {-(x-2)(x-1)}}=(x-2)t}$. Thus

${\displaystyle x={\frac {-2t^{2}-1}{-t^{2}-1}}\qquad \ dx={\frac {2t}{(-t^{2}-1)^{2}}}\,\ dt,}$

and

${\displaystyle {\sqrt {-x^{2}+3x-2}}=(x-2)t={\frac {t}{-t^{2}-1.}}}$

Next,

${\displaystyle \int {\frac {x^{2}}{\sqrt {-x^{2}+3x-2}}}\ dx=\int {\frac {({\frac {-2t^{2}-1}{-t^{2}-1}})^{2}{\frac {2t}{(-t^{2}-1)^{2}}}}{\frac {t}{-t^{2}-1}}}\ dt=\int {\frac {2(-2t^{2}-1)^{2}}{(-t^{2}-1)^{3}}}\ dt.}$

As we can see this is a rational function which can be solved using partial fractions.