Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0

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Theorem

Let $a, b, p, q \in \R$ such that $a p \ne b q$.

Let $a p < 0$.

Then:

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C$

for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.


Proof 1

Let us make the substitution:

\(\text {(1)}: \quad\) \(\ds u\) \(=\) \(\ds \sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } }\) This is valid, because as $a p < 0$ we have $\dfrac {-p} a > 0$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d u} {\d x}\) \(=\) \(\ds \sqrt {\dfrac {-p} a} \map {\dfrac \d {\d x} } {\sqrt {\dfrac {a x + b} {p x + q} } }\) Derivative of Constant Multiple
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {-p} a} \dfrac 1 {2 \sqrt {\dfrac {a x + b} {p x + q} } } \times \map {\dfrac \d {\d x} } {\dfrac {a x + b} {p x + q} }\) Power Rule for Derivatives, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac 1 2 \sqrt {\dfrac {-p} a} \dfrac {\sqrt {p x + q} } {\sqrt {a x + b} } \times \dfrac {\paren {p x + q} \map {\dfrac \d {\d x} } {a x + b} - \paren {a x + b} \map {\dfrac \d {\d x} } {p x + q} } {\paren {p x + q}^2}\) Quotient Rule for Derivatives, some simplification
\(\ds \) \(=\) \(\ds \dfrac 1 2 \sqrt {\dfrac {-p} a} \dfrac {\sqrt {p x + q} } {\sqrt {a x + b} } \times \dfrac {a \paren {p x + q} - p \paren {a x + b} } {\paren {p x + q}^2}\) Derivative of Identity Function, Derivative of Constant Multiple
\(\ds \) \(=\) \(\ds \dfrac 1 2 \sqrt {\dfrac {-p} a} \dfrac {\sqrt {p x + q} } {\sqrt {a x + b} } \times \dfrac {a p x + a q - p a x - p b} {\paren {p x + q}^2}\) mulitplying out
\(\ds \) \(=\) \(\ds \dfrac 1 2 \sqrt {\dfrac {-p} a} \dfrac {\sqrt {p x + q} } {\sqrt {a x + b} } \times \dfrac {a q - b p} {\paren {p x + q}^2}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 2 \sqrt {\dfrac {-p} a} \dfrac 1 {\sqrt {p x + q} \sqrt {a x + b} } \times \dfrac {a q - b p} {p x + q}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d x} {\d u}\) \(=\) \(\ds 2 \sqrt {\dfrac {-a} p} \paren {\sqrt {p x + q} \sqrt {a x + b} } \times \dfrac {p x + q} {a q - b p}\) Derivative of Inverse Function
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \int \frac {\d u} {\sqrt {\paren {a x + b} \paren {p x + q} } } 2 \sqrt {\dfrac {-a} p} \paren {\sqrt {p x + q} \sqrt {a x + b} } \times \dfrac {p x + q} {a q - b p}\) Integration by Substitution
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds 2 \sqrt {\dfrac {-a} p} \int \dfrac {p x + q} {a q - b p} \rd u\) simplifying


Then:

\(\ds u\) \(=\) \(\ds \sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } }\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds u^2\) \(=\) \(\ds \dfrac {-p \paren {a x + b} } {a \paren {p x + q} }\)
\(\ds \leadsto \ \ \) \(\ds u^2 \paren {a \paren {p x + q} }\) \(=\) \(\ds -p \paren {a x + b}\)
\(\ds \leadsto \ \ \) \(\ds u^2 a p x + u^2 a q\) \(=\) \(\ds -p a x - p b\)
\(\ds \leadsto \ \ \) \(\ds u^2 a p x + p a x\) \(=\) \(\ds -u^2 a q - p b\)
\(\ds \leadsto \ \ \) \(\ds a p x \paren {u^2 + 1}\) \(=\) \(\ds -u^2 a q - p b\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac 1 {-a p} \dfrac {a q u^2 + p b} {u^2 + 1}\)


Hence:

\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds 2 \sqrt {\dfrac {-a} p} \int \dfrac {p x + q} {a q - b p} \rd u\) from $(2)$
\(\ds \) \(=\) \(\ds 2 \sqrt {\dfrac {-a} p} \int \dfrac {p \paren {\frac 1 {-a p} \frac {a q u^2 + p b} {u^2 + 1} } + q} {a q - b p} \rd u\) substituting for $x$ from $(3)$
\(\ds \) \(=\) \(\ds 2 \sqrt {\dfrac {-a} p} \int \dfrac {a q u^2 + p b - a q \paren {u^2 + 1} } {-a \paren {u^2 + 1} \paren {a q - b p} } \rd u\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \int \dfrac {a q u^2 + p b - a q u^2 - a q} {\paren {u^2 + 1} \paren {b p - a q} } \rd u\) multiplying out and consolidating constants
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \int \dfrac {b p - a q} {\paren {u^2 + 1} \paren {b p - a q} } \rd u\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \int \dfrac {\d u} {u^2 + 1}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \arctan u + C\) Primitive of Arctangent Function
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C\) substituting for $u$ from $(1)$

$\blacksquare$


Proof 2

First let us express the integrand in the following form:

\(\text {(1)}: \quad\) \(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \int \frac {\d x} {\sqrt {a p \paren {x - \paren {-\frac b a} } \paren {x - \paren {-\frac q p} } } }\)


Recall the definition of Euler's third substitution:

Let $a x^2 + b x + c$ have real roots $\alpha$ and $\beta$.


Euler's third substitution is the substitution:

$\ds \sqrt {a x^2 + b x + c} = \sqrt {a \paren {x - \alpha} \paren {x - \beta} } = \paren {x - \alpha} t$

Then:

$x = \dfrac {a \beta - \alpha t^2} {a - t^2}$

and hence $\d x$ is expressible as a rational function of $x$.


In this context we have:

\(\ds a\) \(\gets\) \(\ds a p\)
\(\ds \alpha\) \(\gets\) \(\ds -\frac b a\)
\(\ds \beta\) \(\gets\) \(\ds -\frac q p\)


Hence we make the substitution:

\(\ds x\) \(=\) \(\ds \dfrac {a p \paren {-\frac q p} - \paren {-\frac b a} t^2} {a p - t^2}\) Lemmata for Euler's Third Substitution: Lemma $1$
\(\ds \) \(=\) \(\ds \dfrac {-a q + \paren {\frac b a} t^2} {a p - t^2}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {a^2 q - b t^2} {a \paren {t^2 - a p} }\) multiplying top and bottom by $-a$


Then we use:

\(\ds \leadsto \ \ \) \(\ds x - \alpha\) \(=\) \(\ds \dfrac {a \paren {\alpha - \beta} } {t^2 - a}\) Lemmata for Euler's Third Substitution: Lemma $2$
\(\ds \) \(=\) \(\ds \dfrac {a p \paren {\paren {-\frac b a} - \paren {-\frac q p} } } {t^2 - a p}\) substituting $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds x - \alpha\) \(=\) \(\ds \dfrac {a q - p b} {t^2 - a p}\) simplifying


Then:

\(\ds \dfrac {\d x} {\d t}\) \(=\) \(\ds \dfrac {2 t a \paren {\beta - \alpha} } {\paren {a - t^2}^2}\) Lemmata for Euler's Third Substitution: Lemma $3$
\(\ds \) \(=\) \(\ds \dfrac {2 t a p \paren {\paren {-\frac q p} - \paren {-\frac b a} } } {\paren {a p - t^2}^2}\) substituting $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d x} {\d t}\) \(=\) \(\ds \dfrac {2 t \paren {b p - a q} } {\paren {t^2 - a p}^2}\) simplifying


Then:

\(\ds t\) \(=\) \(\ds \sqrt {\dfrac {a \paren {x - \beta} } {x - \alpha} }\) Lemmata for Euler's Third Substitution: Lemma $4$
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {a p \paren {x - \paren {-\frac q p} } } {x - \paren {-\frac b a} } }\) substituting $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {a p \paren {x + \frac q p} } {x + \frac b a} }\) rearranging
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {a^2 \paren {p x + q} } {a x + b} }\) rearranging
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds a \sqrt {\dfrac {p x + q} {a x + b} }\) simplifying


Assembling the pieces:

\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \int \frac {\d t} {\paren {\dfrac {a q - p b} {t^2 - a p} } t} \dfrac {2 t \paren {b p - a q} } {\paren {t^2 - a p}^2}\) Integration by Substitution, using $\sqrt {\paren {a x + b} \paren {p x + q} } = \paren {x - \alpha} t$
\(\ds \) \(=\) \(\ds \int \frac {t^2 - a p} {\paren {a q - p b} t} \cdot \dfrac {-2 t \paren {a q - b p} } {\paren {t^2 - a p}^2} \rd t\) simplifying
\(\ds \) \(=\) \(\ds -2 \int \dfrac {\d t} {\paren {t^2 - a p} }\) simplifying
\(\ds \) \(=\) \(\ds -2 \int \dfrac {\d t} {\paren {t^2 + \paren {-a p} } }\) as $a p < 0$ by hypothesis
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arctan {\dfrac t {\sqrt {-a p} } } + C\) Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arctan {\dfrac {a \sqrt {\frac {p x + q} {a x + b} } } {\sqrt {-a p} } } + C\) substituting for $t$ from $(2)$
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-a \paren {p x + q} } {p \paren {a x + b} } } } + C\)
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \map \arccot {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C\) Arccotangent of Reciprocal equals Arctangent
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {-a p} } \paren {\dfrac \pi 2 - \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } } + C\) Sum of Arctangent and Arccotangent
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \paren {\map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } - \dfrac \pi 2} + C\) Arctangent is Odd Function
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C\) subsuming $\dfrac 2 {\sqrt {-a p} } \times \dfrac {-\pi} 2$ into the arbitrary constant

$\blacksquare$


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