Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0/Proof 1
Jump to navigation
Jump to search
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 $(1)$ |
$\blacksquare$