avewright/proofwiki-math huggingface dataset:
Theorem: Reciprocal times Derivative of Gamma Function

Let $z \in \C \setminus \Z_{\le 0}$.
Then:
:$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$
where:
:$\map \Gamma z$ denotes the Gamma function
:$\map {\Gamma'} z$ denotes the derivative of the Gamma function
:$\gamma$ denotes the Euler-Mascheroni constant.

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\map \Gamma z}
      | r = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }
      | c = Weierstrass Form of Gamma Function
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma z
      | r = \frac {e^{-\gamma z} } z \prod_{n \mathop = 1}^\infty \frac {e^{z/n} } {1 + \frac z n}
      | c = reciprocal of both sides
}}
{{eqn | ll= \leadsto
      | l = \map \ln {\map {\Gamma} z}
      | r = \map \ln {\frac {e^{-\gamma z} } z \prod_{n \mathop = 1}^\infty \frac {e^{z/n} } {1 + \frac z n} }
      | c = logarithm of both sides
}}
{{eqn | r = \map \ln {e^{-\gamma z} } - \ln z + \sum_{n \mathop = 1}^\infty \paren { \map \ln {e^{z/n} } -  \map \ln {1 + \frac z n} }
      | c = Sum of Logarithms and Difference of Logarithms
}}
{{eqn | r = -\gamma z \ln e - \ln z + \sum_{n \mathop = 1}^\infty \paren {\frac z n \ln e -  \map \ln {1 + \frac z n} }
      | c = Logarithm of Power
}}
{{eqn | r = -\gamma z - \ln z + \sum_{n \mathop = 1}^\infty \paren {\frac z n - \map \ln {1 + \frac z n} }
      | c = Natural Logarithm of e is 1
}}
{{eqn | ll= \leadsto
      | l = \map {\dfrac \d {\d z} } {\map \ln {\map {\Gamma} z} }
      | r = \map {\dfrac \d {\d z} } {-\gamma z - \ln z + \sum_{n \mathop = 1}^\infty \paren {\frac z n - \map \ln {1 + \frac z n} } }
      | c = differentiating {{WRT|Differentiation}} $z$
}}
{{eqn | ll= \leadsto
      | l = \frac {\map {\Gamma'} z} {\map \Gamma z}
      | r = -\gamma - \frac 1 z + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac {\frac 1 n} {1 + \frac z n} }
      | c = Derivative of Composite Function and Derivative of Natural Logarithm Function
}}
{{eqn | r = -\gamma - \frac 1 z + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n} }
}}
{{eqn | r = -\gamma - \frac 1 z + \paren {\paren {\frac 1 1 - \frac 1 {z + 1} } + \paren {\frac 1 2 - \frac 1 {z + 2} } + \paren {\frac 1 3 - \frac 1 {z + 3} } + \cdots}
}}
{{eqn | r = -\gamma + \paren {\paren {\frac 1 1 - \frac 1 z } + \paren {\frac 1 2 - \frac 1 {z + 1} } + \paren {\frac 1 3 - \frac 1 {z + 2} } + \cdots}
      | c = shifting the terms with $z$ one to the right
}}
{{eqn | r = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Gamma Function at 1

Let $\Gamma$ denote the Gamma function.
Then:
:$\map {\Gamma'} 1 = \ds \int_0^\infty e^{-x} \ln x \rd x = -\gamma$
where:
:$\map {\Gamma'} 1$ denotes the derivative of the Gamma function evaluated at $1$
:$\gamma$ denotes the Euler-Mascheroni constant.

Proof:
From Reciprocal times Derivative of Gamma Function:
:$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$
Setting $z = 1$:
{{begin-eqn}}
{{eqn | l = \frac {\map {\Gamma'} 1} {\map \Gamma 1}
      | r = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {1 + n - 1} }
      | c = 
}}
{{eqn | r = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 n}
      | c = 
}}
{{eqn | r = -\gamma + 0
      | c = 
}}
{{eqn | r = -\gamma
      | c = 
}}
{{end-eqn}}
Using Gamma Function Extends Factorial:
:$\map \Gamma 1 = \paren {1 - 1}! = 1$
Hence:
:$\map {\Gamma'} 1 = -\gamma \map \Gamma 1 = -\gamma$
{{qed}}
Theorem: Commutativity of Parameters of Beta Function

:$\map \Beta {x, y} = \map \Beta {y, x}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Beta {x, y}
      | r = \frac {\map \Gamma x \map \Gamma y} {\map \Gamma {x + y} }
      | c = {{Defof|Beta Function|index = 3}}
}}
{{eqn | r = \frac {\map \Gamma y \map \Gamma x} {\map \Gamma {y + x} }
      | c = Commutative Law of Addition and Commutative Law of Multiplication
}}
{{eqn | r = \map \Beta {y, x}
      | c = {{Defof|Beta Function|index = 3}}
}}
{{end-eqn}}
{{qed}}
Theorem: Beta Function as Integral of Power of t over Power of t plus 1

:$\ds \map \Beta {x, y} = \int_{\mathop \to 0}^{\mathop \to \infty} \frac {t^{x - 1} } {\paren {1 + t}^{x + y} } \rd t$
where $\Beta$ denotes the Beta function.

Proof:
Consider the substitution $s = \dfrac t {1 + t}$.
We have the following:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {1 + t}
      | r = 1 - s
}}
{{eqn | l = t \to 0
      | o = \implies
      | r = s \to 0
}}
{{eqn | l = t \to \infty
      | o = \implies
      | r = s \to 1
}}
{{eqn | l = \d s
      | r = \dfrac 1 {\paren {1 + t}^2} \rd t
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int_{\to 0}^{\to \infty} \frac {t^{x - 1} } {\paren {1 + t}^{x + y} } \rd t
      | r = \int_{\to 0}^{\to \infty} \paren {\frac t {1 + t} }^{x - 1} \paren {\frac 1 {1 + t} }^{y - 1} \frac {\rd t} {\paren {1 + t}^2}
      | c = 
}}
{{eqn | r = \int_{\to 0}^{\to 1} s^{x - 1} \paren {1 - s}^{y - 1} \rd s
      | c = Integration by Substitution
}}
{{eqn | r = \map \Beta {x, y}
      | c = {{Defof|Beta Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Beta Function as Integral of Power of t by Power of 1 minus t over Power of r plus t

:$\ds \map \Beta {x, y} := r^y \paren {r + 1}^x \int_{\mathop \to 0}^{\mathop \to 1} \frac {t^{x - 1} \paren {1 - t}^{y - 1} } {\paren {r + t}^{x + y} } \rd t$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac {\paren {r + 1} t} {\paren {r + t} }
      | c = Integration by Substitution
}}
{{eqn | ll= \leadsto
      | l = \rd u
      | r = \frac {\paren {r + t} \paren {r + 1} - \paren {r + 1} t} {\paren {r + t}^2} \rd t
      | c = Quotient Rule for Derivatives
}}
{{eqn | r = \frac {r \paren {r + 1} } {\paren {r + t}^2} \rd t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1 - u
      | r = 1 - \frac {\paren {r + 1} t} {\paren {r + t} }
      | c = 
}}
{{eqn | r = \paren {\frac {r \paren {1 - t} } {\paren {r + t} } }
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = r^y \paren {r + 1}^x \int_{\mathop \to 0}^{\mathop \to 1} \frac {t^{x - 1} \paren {1 - t}^{y - 1} } {\paren {r + t}^{x + y} } \rd t
      | r = \int_{\mathop \to 0}^{\mathop \to 1} \frac {\paren {r + 1}^x t^{x - 1} } {\paren {r + t}^x} \cdot \frac {r^y \paren {1 - t}^{y - 1} } {\paren {r + t}^y} \rd t
      | c = 
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} \frac {\paren {r + 1}^{x - 1} t^{x - 1} } {\paren {r + t}^{x - 1} } \cdot \frac {r^{y - 1} \paren {1 - t}^{y - 1} } {\paren {r + t}^{y - 1} } \cdot \paren {\frac {\paren {r + 1} } {\paren {r + t} } } \cdot \paren {\frac r {\paren {r + t} } } \rd t
      | c = Product of Powers
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} \paren {\frac {\paren {r + 1} t} {\paren {r + t} } }^{x - 1} \cdot \paren {\frac {r \paren {1 - t} } {\paren {r + t} } }^{y - 1} \cdot \paren {\frac {\paren {r + 1} } {\paren {r + t} } } \cdot \paren {\frac r {\paren {r + t} } } \rd t
      | c = Power of Product
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} u^{x - 1} \paren {1 - u}^{y - 1} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \map \Beta {x, y}
      | c = {{Defof|Beta Function|index = 1}}
}}
{{end-eqn}}
{{qed}}
Theorem: Solution to Separable Differential Equation/General Result

Consider the separable differential equation:
:$\map {g_1} x \map {h_1} y + \map {g_2} x \map {h_2} y \dfrac {\d y} {\d x} = 0$
Its general solution is found by solving the integration:
:$\ds \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y = C$

Proof:
{{begin-eqn}}
{{eqn | l = \map {g_1} x \map {h_1} y + \map {g_2} x \map {h_2} y \frac {\d y} {\d x}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x}
      | r = 0
      | c = dividing both sides by $\map {g_2} x \map {h_1} y$
}}
{{eqn | ll= \leadsto
      | l = \int \left({\frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} }\right) \rd x
      | r = \int 0 \rd x
      | c = integrating both sides {{WRT|Integration}} $x$
}}
{{eqn | ll= \leadsto
      | l = \int \left({\frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} }\right) \rd x
      | r = C
      | c = Primitive of Constant
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} \rd x
      | r = C
      | c = Linear Combination of Primitives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y
      | r = C
      | c = Integration by Substitution
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Arithmetic-Geometric Sequence

Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:
:$a_k = \paren {a + k d} r^k$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k
      | r = a \sum_{k \mathop = 0}^{n - 1} r^k + d \sum_{k \mathop = 0}^{n - 1} k r^k
      | c = 
}}
{{eqn | r = \frac {a \paren {1 - r^n} } {1 - r} + d \sum_{k \mathop = 0}^{n - 1} k r^k
      | c = Sum of Geometric Sequence
}}
{{eqn | r = \frac {a \paren {1 - r^n} } {1 - r} + d \paren {\frac {\paren {n - 1} r^{n + 1} - n r^n + r} {\paren {r - 1}^2} }
      | c = Sum of Sequence of Power by Index
}}
{{end-eqn}}
Hence the result, after algebra.
{{qed}}
Theorem: Sum of Infinite Arithmetic-Geometric Sequence

Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:
:$a_k = \paren {a + k d} r^k$ for $n = 0, 1, 2, \ldots$
Let:
:$\size r < 1$
where $\size r$ denotes the absolute value of $r$.
Then:
:$\ds \sum_{n \mathop = 0}^\infty \paren {a + k d} r^k = \frac a {1 - r} + \frac {r d} {\paren {1 - r}^2}$

Proof:
From Sum of Arithmetic-Geometric Sequence, we have:
:$\ds s_n = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$
We have that $\size r < 1$.
So by Sequence of Powers of Number less than One:
:$r^n \to 0$ as $n \to \infty$
and
:$r^{n - 1} \to 0$ as $n - 1 \to \infty$
Hence:
:$s_n \to \dfrac a {1 - r} + \dfrac {r d} {\paren {1 - r}^2}$
as $n \to \infty$.
The result follows.
{{qed}}
Theorem: Reduction Formula for Integral of Power of Cosine

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Let:
:$I_n := \ds \int \cos^n x \rd x$
Then:
:$I_n = \dfrac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n I_{n - 2}$
is a reduction formula for $\ds \int \cos^n x \rd x$.

Proof:
With a view to expressing the problem in the form:
:$\ds \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \cos^{n - 1} x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\rd u} {\rd x}
      | r = -\paren {n - 1} \cos ^{n - 2} x \sin x
      | c = Chain Rule for Derivatives, Derivative of Cosine Function, Derivative of Power
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\rd v} {\rd x}
      | r = \cos x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sin x
      | c = Primitive of Cosine Function
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = I_n
      | r = \int \cos^n x \rd x
      | c = 
}}
{{eqn | r = \int \cos^{n - 1} x \cos x \rd x
      | c = 
}}
{{eqn | r = \cos^{n - 1} x \sin x - \int \sin x \paren {-\paren {n - 1} \cos^{n - 2} x \sin x} \rd x
      | c = Integration by Parts
}}
{{eqn | r = \cos^{n - 1} x \sin x + \int \paren {n - 1} \cos^{n - 2} x \sin^2 x \rd x
      | c = rearranging
}}
{{eqn | r = \cos^{n - 1} x \sin x + \int \paren {n - 1} \cos^{n - 2} x \paren {1 - \cos^2 x} \rd x
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \cos^{n - 1} x \sin x + \paren {n - 1} \int \cos^{n - 2} x \rd x - \int \paren {n - 1} \cos^n x \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \cos^{n - 1} x \sin x + \paren {n - 1} I_{n - 2} - \paren {n - 1} I_n
      | c = substituting for $I_n$ and $I_{n - 2}$
}}
{{eqn | ll= \leadsto
      | l = n I_n
      | r = \cos^{n - 1} x \sin x + \paren {n - 1} I_{n - 2}
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = I_n
      | r = \frac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n I_{n - 2}
      | c = dividing both sides by $n$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Root of a x squared plus b x plus c/a equal to 0

:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {2 \sqrt {b x + c} } b + C$
when $a = 0$.

Proof:
{{begin-eqn}}
{{eqn | l = a
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | r = \int \frac {\d x} {\sqrt {b x + c} }
      | c = 
}}
{{eqn | r = \frac {2 \sqrt {b x + c} } b
      | c = Primitive of $\dfrac 1 {\sqrt {a x + b} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c/b equal to 0/Proof 1

Let $a \in \R_{\ne 0}$.
{{:Primitive of Reciprocal of a x squared plus b x plus c/b equal to 0}}

Proof:
First:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {\d x} {a x^2 + c}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {\d x} {x^2 + \frac c a}
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Let $a c > 0$.
Then $\dfrac c a > 0$ and:
{{begin-eqn}}
{{eqn | l = \frac c a
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac c a
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {\frac c a}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \frac 1 a \int \frac {\d x} {x^2 + q^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a q} \arctan \frac x q + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \frac 1 {a \sqrt {\frac c a} } \arctan \frac x {\sqrt {\frac c a} } + C
      | c = substituting for $q$
}}
{{eqn | r = \dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $a c < 0$.
Then $\dfrac c a > 0$ and:
{{begin-eqn}}
{{eqn | l = \frac c a
      | o = <
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\frac c a
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {-\frac c a}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \frac 1 a \int \frac {\d x} {x^2 - q^2}
      | c = 
}}
{{eqn | r = \frac 1 {2 a q} \ln \size {\frac {z - q} {z + q} } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {2 a \sqrt {-\frac c a} } \ln \size {\frac {z - \sqrt {-\frac c a} } {z + \sqrt {-\frac c a} } } + C
      | c = substituting for $q$
}}
{{eqn | r = \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $c = 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {\d x} {a x^2}
      | c =
}}
{{eqn | r = \dfrac {-1} {a x} + C
      | c = Primitive of Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} } = \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$

Proof:
First:
{{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
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} }
      | r = \frac 1 {2 a} \int \frac {2 a x \rd x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\sqrt {a x^2 + b x + c} } - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } 
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\d z} {\sqrt z} - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } 
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {2 a} 2 \sqrt z - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = Primitive of Power
}}
{{eqn | r = \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Root of a x squared plus b x plus c

Let $a, b, c \in \R_{\ne 0}$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$ and $x \ne 0$:
$\quad \ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } = \begin {cases}
\dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x} + C & : c > 0, b^2 - 4 a c > 0 \\
\dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {\size x \sqrt {4 a c - b^2} } } + C & : c > 0, b^2 - 4 a c < 0 \\
\dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 c} x + b} + C & : c > 0, b^2 - 4 a c = 0 \\
\dfrac 1 {\sqrt {-c} } \map \arcsin {\dfrac {b x + 2 c} {\size x \sqrt {\size {b^2 - 4 a c} } } } & : c < 0, b^2 - 4 a c \ne 0  \\
\end {cases}$

Proof:
{{begin-eqn}}
{{eqn | l = x \sqrt {a x^2 + b x + c}
      | r = \frac x {\paren {a x^2 + b x + c}^{-\frac 1 2} }
      | c = 
}}
{{eqn | r = \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {\paren {a x^2 + b x + c}^{-\frac 1 2} \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} }
      | c = 
}}
{{eqn | r = \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} }
      | c = 
}}
{{eqn | r = \frac N D
      | c = 
}}
{{eqn-intertext | where:}}
{{eqn | l = N
      | r = \frac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x
      | c = 
}}
{{eqn | l = D
      | r = \frac {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac {\d N} {\d x}
      | r = \frac {x \paren {\sqrt c \paren {2 a x + b} \paren {a x^2 + b x + c}^{-\frac 1 2} + b} - \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {x^2}
      | c = 
}}
{{eqn | r = \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x} + b x \sqrt {a x^2 + b x + c} } - \paren {2 \sqrt c \paren {a x^2 + b x + c} + \paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}
      | c = 
}}
{{eqn | r = \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x - 2 a x^2 - 2 b x - 2 c} + b x \sqrt {a x^2 + b x + c} } - \paren {\paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}
      | c = 
}}
{{eqn | r = \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {-b x - 2 c} - \paren {\paren {2 c} \sqrt {a x^2 + b x + c} } } } {x^2}
      | c = 
}}
{{eqn | r = -\frac {2 c + \sqrt c \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}
      | c = 
}}
{{eqn | r = \sqrt c \times D
      | c = 
}}
{{end-eqn}}
Now provided $N$ is real and non zero:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | r = \int \frac D N \rd x
      | c = 
}}
{{eqn | r = \frac {-1} {\sqrt c} \int \frac {\frac {\d N} {\d x} } N \rd x
      | c = 
}}
{{eqn | r = \frac {-1} {\sqrt c} \ln \size N
      | c = Primitive of Function under its Derivative
}}
{{eqn | r = \frac {-1} {\sqrt c} \ln \size {\frac {\paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } x}
      | c = 
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal of x squared by Root of a x squared plus b x plus c

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

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\d x} {x^2 \sqrt {a x^2 + b x + c} }
}}
{{eqn | r = \int \frac {c \d x} {c x^2 \sqrt {a x^2 + b x + c} }
      | c = multiplying top and bottom by $c$
}}
{{eqn | r = \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x^2 \sqrt {a x^2 + b x + c} }
      | c = adding and subtracting $a x^2 + b x$
}}
{{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x^2 \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {x^2 \rd x} {x^2 \sqrt {a x^2 + b x + c} } - \frac b c \int \frac {x \rd x} {x^2 \sqrt {a x^2 + b x + c} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 c \int \frac {\sqrt {a x^2 + b x + c} \rd x} {x^2} - \frac a c \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac b c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = simplification
}}
{{eqn | r = \frac 1 c \paren {-\frac {\sqrt {a x^2 + b x + c} } x + a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + \frac b 2 \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\dfrac {\sqrt {a x^2 + b x + c} } {x^2}$
}}
{{eqn | o = 
      | ro= -
      | r = \frac a c \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac b c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = -\frac {\sqrt {a x^2 + b x + c} } {c x} - \frac b {2 c} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int x \sqrt {a x^2 + b x + c} \rd x = \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac {b \paren {2 a x + b} \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \paren {4 a c - b^2} } {16 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int x \sqrt {a x^2 + b x + c} \rd x
}}
{{eqn | r = \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x
      | c = Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$
}}
{{eqn | r = \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a}
      | c =
}}
{{eqn | o = 
      | ro= -
      | r = \frac b {2 a} \paren {\frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\sqrt {a x^2 + b x + c}$
}}
{{eqn | r = \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a}
      | c =
}}
{{eqn | o = 
      | ro= -
      | r = \frac {b \paren {2 a x + b} \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \paren {4 a c - b^2} } {16 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}