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

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Theorem

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

Let $a p > 0$.

Then:

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

for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.


Proof

Lemma $1$

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

Then:

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

$\Box$


Lemma $2$

$\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:

$u := \sqrt {a x + b}$

$\Box$


We have by hypothesis that $a p > 0$.

Thus:

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


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

Then we set:

$(2): \quad c^2 :=\dfrac {b p - a q} p$


Then:

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

$\Box$


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

Then we set:

$(3): \quad c^2 := -\dfrac {b p - a q} p$


Then:

\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 + c^2} }\) from $(3)$
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a p} } \ln \size {\sqrt {a x + b} + \sqrt {u^2 + c^2} } + C\) Primitive of $\dfrac 1 {\sqrt {u^2 + c^2} }$: logarithm form
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a p} } \ln \size {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C\) substituting for $u$ and $c$


Thus we have in both cases that:

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


Then:

\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\ds \frac 2 {\sqrt {a p} } \ln \size {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C\)
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a p} } \ln \size {\sqrt {a x + b} + \sqrt {\frac a p} \sqrt {p x + q} } + C\) Lemma $1$
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a p} } \ln \size {\frac {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } {\sqrt p} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } - \ln {\sqrt p} + C\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C\) subsuming $-\ln {\sqrt p}$ into the arbitrary constant

$\blacksquare$


Sources

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