Primitive of Reciprocal of Root of a minus x by Cube of Root of x minus b/Proof 2

Theorem

$\ds \int \dfrac {\d x} {\paren {a - x}^{1/2} \paren {x - b}^{3/2} } = \dfrac 2 {b - a} \sqrt {\dfrac {a - x} {x - b} } + C$

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$

$=$

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

$\ds \leadsto \ \ $

$\ds \int \dfrac {\d x} {\paren {a - x}^{1/2} \paren {x - b}^{3/2} }$

$=$

$\ds \frac {2 \sqrt {\paren {-1} x + a} } {\paren {\paren {-1} \paren {-b} - a \cdot 1} \sqrt {1 \cdot x + \paren {-b} } } + C$

setting $p \gets 1$, $q \gets b$, $a \gets -1$, $b \gets a$

$\ds $

$=$

$\ds \frac {2 \sqrt {a - x} } {\paren {b - a} \sqrt {x - b} } + C$

simplifying

$\ds $

$=$

$\ds \dfrac 2 {b - a} \sqrt {\dfrac {a - x} {x - b} } + C$

rearranging

$\blacksquare$