Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) greater than 0/Also presented as
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Primitive of $\frac 1 {\paren {p x + q} \sqrt {a x + b} }$ where $p \paren {b p - a q} > 0$: 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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.114$