Theorem: Primitive of Reciprocal of x by Half Integer Power of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} } = \frac 1 {\paren {2 n - 1} c \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }$

Proof:
{{begin-eqn}}
{{eqn | r = \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | o = 
      | c = 
}}
{{eqn | r = \int \frac {c \rd x} {c x \paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = multiplying top and bottom by $c$
}}
{{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x \paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = 
}}
{{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac a c \int \frac {x^2 \rd x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b c \int \frac {x \rd x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = Linear Combination of Primitives
}}
{{eqn | n = 1
      | r = \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac a c \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = simplifying
}}
{{end-eqn}}
Now take:
{{begin-eqn}}
{{eqn | r = \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | o = 
      | c = 
}}
{{eqn | r = \int \frac {2 a x \rd x} {2 a \paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = multiplying top and bottom by $2 a$
}}
{{eqn | r = \int \frac {\paren {2 a x + b - b} \rd x} {2 a \paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = 
}}
{{eqn | n = 2
      | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a x^2 + b x + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x} 
      | r = 2 a x + b
      | c = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | r = \int z^{-n - \frac 1 2} \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac {z^{-n + \frac 1 2} } {-n + \frac 1 2}
      | c = Primitive of Power
}}
{{eqn | n = 3
      | r = \frac {-2} {\paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} }
      | c = simplifying
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | r = \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | o = 
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = from $(2)$
}}
{{eqn | r = \frac 1 {2 a} \paren {\frac {-2} {\paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} } } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = from $(3)$
}}
{{eqn | r = \frac {-1} {a \paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = simplifying
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | r = \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | o = 
      | c = 
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac a c \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = from $(1)$
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac a c \paren {\frac {-1} {a \paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } }
      | c = from $(3)$
}}
{{eqn | o = 
      | ro= -
      | r = \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = 
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} }
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \frac 1 {\paren {2 n - 1} c \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = simplifying
}}
{{eqn | o = 
      | ro= -
      | r = \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = 
}}
{{eqn | r = \frac 1 {\paren {2 n - 1} c \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | c = gathering terms
}}
{{end-eqn}}
{{qed}}