Primitive of Reciprocal of Root of a x squared plus b x plus c/a equal to 0

Theorem

For $a x^2 + b x + c > 0$:

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

when $a = 0$.

Proof

\(\ds a\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \int \frac {\d x} {\sqrt {b x + c} }\)

\(\ds \)

\(=\)

\(\ds \frac {2 \sqrt {b x + c} } b\)

Primitive of $\dfrac 1 {\sqrt {a x + b} }$

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.280$