Primitive of Reciprocal of Root of a x squared plus b x plus c/a greater than 0/Positive Discriminant

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Theorem

Let $a, b, c \in \R$ such that $a > 0$.

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

Then for $x \in \R$ such that $a x^2 + b x + c > 0$:

$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C$


Proof

Completing the Square

First:

\(\ds a x^2 + b x + c\) \(=\) \(\ds \frac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}\) Completing the Square
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }\)


Put:

\(\ds z\) \(=\) \(\ds 2 a x + b\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a\) Derivative of Power


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

Thus:

\(\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }\) from $(1)$
\(\ds \) \(=\) \(\ds \int \frac {\d z} {\sqrt a \sqrt {z^2 - D} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - D} }\) Primitive of Constant Multiple of Function

$\Box$


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

Then:

\(\ds D\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds D\) \(=\) \(\ds q^2\) for some $q \in \R$
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds \sqrt {b^2 - 4 a c}\) by definition of $D$


Thus:

\(\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - q^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {z + \sqrt {z^2 - q^2} } + C\) Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {2 a x + b + \sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } } + C\) substituting for $z$ and $q$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {2 a x + b + \sqrt {4 a \paren {a x^2 + b x + c} } } + C\) Completing the Square
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C\) rearranging

$\blacksquare$


Sources

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