Primitive of Reciprocal of Root of a x + b by Root of p x + q/Completing the Square

From ProofWiki
Jump to navigation Jump to search

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

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_Root_of_a_x_%2B_b_by_Root_of_p_x_%2B_q/Completing_the_Square&oldid=631081"