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

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Theorem

Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.

Let $p \paren {b p - a q} > 0$.

Then:

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


Proof

Lemma

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

where:

$u = \sqrt {a x + b}$

$\Box$


We have by hypothesis that:

$p \paren {b p - a q} < 0$

which means:

$\dfrac {b p - a q} p < 0$

Hence let:

$d^2 = -\dfrac {b p - a q} p$

Thus:

\(\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }\) \(=\) \(\ds \frac 2 p \int \frac {\d u} {u^2 + \paren {-\dfrac {b p - a q} p} }\) Lemma
\(\ds \) \(=\) \(\ds \frac 2 p \int \frac {\d u} {u^2 + d^2}\) setting $d^2 = \dfrac {b p - a q} p$
\(\ds \) \(=\) \(\ds \frac 2 p \frac 1 d \arctan {\frac u d} + C\) Primitive of $\dfrac 1 {u^2 + d^2}$
\(\ds \) \(=\) \(\ds \frac 2 p \sqrt {-\dfrac p {b p - a q} } \arctan {\sqrt {-\dfrac p {b p - a q} } \sqrt {a x + b} } + C\) substituting for $d$ and $u$
\(\ds \) \(=\) \(\ds \dfrac {-2} {\sqrt {p \paren {a q - b p} } } \arctan {-\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan {\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C\) Arctangent Function is Odd

$\blacksquare$


Also presented as

This result can also be seen presented as:

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

but this presupposes both that $p > 0$ and $b p - a q < 0$.


Sources

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