Primitive of Reciprocal of Root of a x + b by Root of p x + q/Completing the Square
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Lemma for Primitive of $\frac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
Then:
- $\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\ds \frac 1 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {b p - a q}^2} } & : a p > 0 \\ \ds \frac 1 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {b p - a q}^2 - u^2} } & : a p < 0 \end {cases}$
where:
- $u := 2 a p x + b p + a q$
Proof
\(\ds \paren {a x + b} \paren {p x + q}\) | \(=\) | \(\ds a p x^2 + \paren {b p + a q} + b q\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a p x + b p + a q}^2 + 4 a p b q - \paren {b p + a q}^2} {4 a p}\) | Completing the Square: $a \gets a p$, $b \gets \paren {b p + a q}$, $c \gets b q$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a p x + b p + a q}^2 + 4 a p b q - \paren {\paren {b p}^2 + 2 a p b q + \paren {a q}^2} } {4 a p}\) | Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a p x + b p + a q}^2 - \paren {\paren {b p}^2 - 2 a p b q + \paren {a q}^2} } {4 a p}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a p x + b p + a q}^2 - \paren {b p - a q}^2} {4 a p}\) | Square of Difference | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) | \(=\) | \(\ds \int \sqrt {\frac {4 a p} {\paren {2 a p x + b p + a q}^2 - \paren {b p - a q}^2} } \rd x\) |
Let us make the substitution:
\(\ds u\) | \(=\) | \(\ds 2 a p x + b p + a q\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac {\d u} {\d x}\) | \(=\) | \(\ds 2 a p\) |
- Case $1
- \quad a p > 0$
\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) | \(=\) | \(\ds \int \sqrt {\frac {4 a p} {\paren {2 a p x + b p + a q}^2 - \paren {b p - a q}^2} } \rd x\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac 1 {2 a p} \sqrt {\frac {4 a p} {u^2 - \paren {b p - a q}^2} } \rd u\) | Integration by Substitution: from $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {b p - a q}^2} }\) | simplification |
$\Box$
- Case $2
- \quad a p < 0$
\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) | \(=\) | \(\ds \int \sqrt {\frac {4 a p} {\paren {2 a p x + b p + a q}^2 - \paren {b p - a q}^2} } \rd x\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \sqrt {\frac {-4 \paren {-a p} } {\paren {2 a p x + b p + a q}^2 - \paren {b p - a q}^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \sqrt {\frac {4 \paren {-a p} } {\paren {b p - a q}^2 - \paren {2 a p x + b p + a q}^2} } \rd x\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac 1 {2 a p} \sqrt {\frac {4 \paren {-a p} } {\paren {b p - a q}^2 - u^2} } \rd u\) | Integration by Substitution: from $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {b p - a q}^2 - u^2} }\) | simplification |
$\blacksquare$