Primitive of Reciprocal of Root of a x + b by Root of p x + q/Lemma 2
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Lemma for Primitive of $\frac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$
- $\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\ds \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } } & : a p > 0 \\ \ds \frac 2 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {\frac {b p - a q} p} - u^2} } & : a p < 0 \end {cases}$
where:
- $u := \sqrt {a x + b}$
Proof
Let us make the substitution:
- $u = \sqrt {a x + b}$
Lemma
Let $u = \sqrt {a x + b}$.
Then:
- $\ds \sqrt {p x + q} = \sqrt {\paren {\frac p a} \paren {u^2 - \paren {\frac {b p - a q} p} } }$
$\Box$
- Case $1
- \quad a p > 0$
\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) | \(=\) | \(\ds \int \frac {2 u \rd u} {a \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} } u}\) | Primitive of Function of $\sqrt {p x + q}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } }\) | Primitive of Constant Multiple of Function |
$\Box$
- Case $2
- \quad a p < 0$
We have:
\(\ds a p\) | \(<\) | \(\ds 0\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac p a\) | \(<\) | \(\ds 0\) |
Then:
\(\ds \sqrt {p x + q}\) | \(=\) | \(\ds \sqrt {\paren {\frac p a} \paren {u^2 - \paren {\frac {b p - a q} p} } }\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\paren {-\frac p a} \paren {\paren {\frac {b p - a q} p} - u^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {-\frac p a} \sqrt {\paren {\frac {b p - a q} p} - u^2}\) |
Then:
\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) | \(=\) | \(\ds \dfrac 2 p \int \frac {u \rd u} {\sqrt {-\frac p a} \paren {\sqrt {\paren {\frac {b p - a q} p} - u^2} } u}\) | Primitive of Function of $\sqrt {p x + q}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {\frac {b p - a q} p} - u^2} }\) | Primitive of Constant Multiple of Function and simplifying |
$\blacksquare$