Lemmata for Euler's Third Substitution/Lemma 2

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:

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

$\ds x$

$=$

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

$\ds \leadsto \ \ $

$\ds x - \alpha$

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

simplifying

$\blacksquare$