Lemmata for Euler's Third Substitution/Lemma 3

Theorem

Let Euler's third substitution be employed in order to evaluate a primitive of the form:

$\ds \map R {x, \sqrt {a x^2 + b x + c} } \rd x$

where $R$ is a rational function of $x$ and $\sqrt {a x^2 + b x + c}$.

Thus:

$\ds \sqrt {a x^2 + b x + c} = \sqrt {a \paren {x - \alpha} \paren {x - \beta} } = \paren {x - \alpha} t$

Then we have:

$\dfrac {\d x} {\d t} = \dfrac {2 t a \paren {\beta - \alpha} } {\paren {a - t^2}^2}$

$\ds \dfrac {\d x} {\d t}$

$=$

$\ds \map {\dfrac \d {\d t} } {x - \alpha}$

$\ds $

$=$

$\ds \map {\dfrac \d {\d t} } {\dfrac {a \paren {\alpha - \beta} } {t^2 - a} }$

$\ds $

$=$

$\ds \dfrac {-a \paren {\alpha - \beta} } {\paren {t^2 - a}^2} \cdot \paren {2 t}$

$\ds \leadsto \ \ $

$\ds \dfrac {\d x} {\d t}$

$=$

$\ds \dfrac {2 t a \paren {\beta - \alpha} } {\paren {a - t^2}^2}$

simplifying

$\blacksquare$