Primitive of Reciprocal of Root of a x squared plus b x plus c/a less than 0/Zero Discriminant

Theorem

Let $a \in \R_{\ne 0}$.

Let $b^2 - 4 a c = 0$.

Then:

$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$

is not defined.

Proof

Suppose that $b^2 - 4 a c = 0$.

Then:

\(\ds a x^2 + b x + c\)

\(=\)

\(\ds \frac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}\)

Completing the Square

\(\ds \)

\(=\)

\(\ds \frac {\paren {2 a x + b}^2} {4 a}\)

as $b^2 - 4 a c = 0$

But we have that:

$\paren {2 a x + b}^2 > 0$

while under our assertion that $a < 0$:

$4 a < 0$

and so:

$a x^2 + b x + c < 0$

Thus on the real numbers $\sqrt {a x^2 + b x + c}$ is not defined.

Hence it follows that:

$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$

is not defined.

$\blacksquare$