Primitive of Reciprocal of p x + q by Root of a x + b

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Theorem

$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases} \dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\ \end {cases}$


Proof


Let:

\(\displaystyle u\) \(=\) \(\displaystyle \sqrt {a x + b}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \frac {u^2 - b} a\)


Then:

\(\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }\) \(=\) \(\displaystyle \int \frac {2 u} {p \paren {\paren {\dfrac {u^2 - b} a} + q} a u} \rd u\) Primitive of Function of $\sqrt{a x + b}$
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {2 a u} {\paren {p \paren {u^2 - b} + a q} a u} \rd u\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle 2 \int \frac {\d u} {p u^2 - b p + a q}\) simplifying
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}\) putting denominator into the form $z^2 - a^2$


Now let $z = \sqrt p u$

\(\displaystyle z\) \(=\) \(\displaystyle \sqrt p u\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d z} {\d u}\) \(=\) \(\displaystyle \sqrt p\)


Thus:

\(\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }\) \(=\) \(\displaystyle 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}\) from $(1)$
\(\text {(2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt p} \int \frac {\d z} {z^2 - \paren {\sqrt {b p - a q} }^2}\) simplifying


Let $b p - a q > 0$.

Let $d = \sqrt {b p - a q}$.

Then:

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

$\Box$


Let $b p - a q < 0$.

Let $d = \sqrt {-\paren {b p - a q} } = \sqrt {a q - b p}$.

Then:

\(\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }\) \(=\) \(\displaystyle \frac 2 {\sqrt p} \int \frac {\d z} {z^2 + d^2}\) from $(2)$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt p} \frac 1 d \arctan {\frac z d} + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt p d} \arctan {\frac {\sqrt p u} d} + C\) substituting for $z$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a q - b p} \sqrt p} \arctan {\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C\) substituting for $u$ and $d$

$\blacksquare$


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