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

From ProofWiki
Jump to navigation Jump to search

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:

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \frac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$


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 $d^2 = \dfrac {b p - a q} p$
\(\ds \) \(=\) \(\ds \frac 2 p \frac 1 {2 d} \ln \size {\frac {u - d} {u + d} } + C\) Primitive of $\dfrac 1 {u^2 - d^2}$: Logarithm Form
\(\ds \) \(=\) \(\ds \frac 2 p \dfrac 1 2 \sqrt {\dfrac p {b p - a q} } \ln \size {\frac {u - \sqrt {\dfrac {b p - a q} p } } {u + \sqrt {\dfrac {b p - a q} p} } } + C\) substituting for $d$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\frac {\sqrt {p u} - \sqrt {b p - a q} } {\sqrt {p u} + \sqrt {b p - a q} } } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C\) substituting for $u$

$\blacksquare$


Also presented as

This result can also be seen presented as:

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \frac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$

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


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_p_x_%2B_q_by_Root_of_a_x_%2B_b/p_(b_p_-_a_q)_greater_than_0&oldid=631106"