Primitive of Reciprocal of x by Root of a x squared plus b x plus c/Lemma

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Theorem

Let $a, b, c \in \R_{\ne 0}$.

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

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

where $u := \dfrac 1 x$, according to whether $u > 0$ or $u < 0$.


Proof

\(\ds x\) \(=\) \(\ds \frac 1 u\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d u}\) \(=\) \(\ds \frac {-1} {u^2}\) Primitive of Power
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \int {\frac 1 {\frac 1 u \sqrt {a \paren {\frac 1 u}^2 + b \paren {\frac 1 u} + c} } } \frac {-\rd u} {u^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\int \frac {\d u} {\frac {u^2} u \sqrt {\frac 1 {u^2} \paren {a + b u + c u^2} } }\) Primitive of Constant Multiple of Function and simplification
\(\ds \) \(=\) \(\ds -\int \frac {\d u} {\frac {u^2} u \sqrt {\frac 1 {u^2} } \sqrt {a + b u + c u^2} }\) extracting $\sqrt {\frac 1 {u^2} }$
\(\ds \) \(=\) \(\ds -\int \frac {\d u} {\pm \sqrt {a + b u + c u^2} }\) according to whether $u > 0$ or $u < 0$

$\blacksquare$


Sources

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