Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0/Proof 1

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

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 $(1)$

$\blacksquare$