avewright/proofwiki-math huggingface dataset:
Theorem: Fekete's Subadditive Lemma

Let $\sequence {a_n}_{n \mathop \ge 1}$ be a subadditive sequence.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {a_n} n = \inf_{n \mathop \ge 1} \frac {a_n} n$

Proof:
Let $k \ge 1$.
Let $n \ge 1$.
By Division Theorem, there exist $q \in \N$ and $r \in \set {0, 1, \ldots , k - 1}$ such that:
:$n = k q + r$
Thus:
{{begin-eqn}}
{{eqn | l = \frac {a_n} n
      | r = \frac {a_{k q + r} } n
}}
{{eqn | o = \le
      | r = \frac {a_{k q} + a_r} n
      | c = {{Defof|Subadditive Sequence}}
}}
{{eqn | r = \frac {a_{\underbrace {k + \cdots + k}_q} + a_r} n
}}
{{eqn | o = \le
      | r = \frac {q a_k + a_r} n
      | c = {{Defof|Subadditive Sequence}}
}}
{{eqn | r = \frac {\frac {n - r} k a_k + a_r} n
}}
{{eqn | r = \frac {a_k} k + \frac {a_r - \frac r k a_k} n
}}
{{eqn | o = \le
      | r = \frac {a_k} k + \frac {2 \max \set {\size {a_0}, \size {a_1}, \ldots , \size {a_k} } } n
}}
{{end-eqn}}
By $n \to \infty$, we obtain:
:$\ds \forall k \ge 1 : \limsup_{n \mathop \to \infty} \frac {a_n} n \le \frac {a_k} k$
In particular:
:$\ds (1): \quad \limsup_{n \mathop \to \infty} \frac {a_n} n \le \inf_{k \mathop \ge 1} \frac {a_k} k$
On the other hand, by definition of infimum:
:$\ds \forall n \ge 1 : \inf_{k \mathop \ge 1} \frac {a_k} k \le \frac {a_n} n$
In particular:
:$\ds (2): \quad \inf_{k \mathop \ge 1} \frac {a_k} k \le \liminf_{n \mathop \to \infty} \frac {a_n} n$
By Convergence of Limsup and Liminf, $(1)$ and $(2)$ together mean:
:$\ds \lim_{n \mathop \to \infty} \frac {a_n} n = \inf_{k \mathop \ge 1} \frac {a_k} k$
{{qed}}
Theorem: Supremum Operator Norm on Continuous Linear Transformation Space is Submultiplicative

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$, $\struct {Z, \norm {\, \cdot \,}_Z}$ be normed vector spaces.
Let $A : Y \to Z$ and $B : X \to Y$ be continuous linear transformations.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $\circ$ denote the composition.
Then $\norm {\, \cdot \,}$ is submultiplicative:
:$\norm {A \circ B} \le \norm A \cdot \norm B$

Proof:
{{begin-eqn}}
{{eqn | l = \norm {A \circ B}
      | r = \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm {\map {\paren {A \circ B} } x}_Z
      | c = {{Defof|Supremum Operator Norm}}
}}
{{eqn | o = \le 
      | r = \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm A \cdot \norm {\map B x}_Y
      | c = Supremum Operator Norm as Universal Upper Bound
}}
{{eqn | r = \norm A \cdot \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm {\map B x}_Y
      | c = $\norm A$ is independent of $x$
}}
{{eqn | o = \le
      | r = \norm A \cdot \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm B \cdot \norm x_X
      | c = Supremum Operator Norm as Universal Upper Bound
}}
{{eqn | r = \norm A \cdot \norm B \cdot \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm x_X
      | c = $\norm B$ is independent of $x$
}}
{{eqn | o = \le
      | r = \norm A \cdot \norm B \cdot 1
}}
{{eqn | r = \norm A \cdot \norm B
}}
{{end-eqn}}
{{qed}}
Theorem: Characterization of Paracompactness in T3 Space/Lemma 14

:$\forall n \in \N_{>0}: U_n \subseteq V_0$

Proof:
=== Proof: $V_n \subseteq V_{n - 1}$ ===
We have:
{{begin-eqn}}
{{eqn | q = \forall n \in \N_{> 0}
      | l = \tuple{x, y} \in V_n
      | o = \leadsto
      | r = \tuple{x, y}, \tuple{y, y} \in V_n
      | c = as $\Delta_X \subseteq V_n$ 
}}
{{eqn | o = \leadsto
      | r = \tuple{x, y} \in V_n \circ V_n
      | c = {{Defof|Composite Relation}}
}}
{{eqn | ll = \leadsto
      | q = \forall n \in \N_{> 0}
      | l = V_n
      | o = \subseteq
      | r = V_n \circ V_n
      | c = {{Defof|Subset}}
}}
{{eqn | q = \forall n \in \N_{> 0}
      | l = V_n \circ V_n
      | o = \subseteq
      | r = V_{n - 1}
      | c = {{hypothesis}}
}}
{{eqn | n = 1
      | ll = \leadsto
      | q = \forall n \in \N_{> 0}
      | l = V_n
      | o = \subseteq
      | r = V_{n - 1}
      | c = From Subset Relation is Transitive
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Existence of Chebyshev Polynomials of the Second Kind

There exists a '''Chebyshev polynomial of the second kind''' for all natural numbers $n$.

Proof:
{{tidy}}
For $n = 0$:
{{begin-eqn}}
{{eqn | l = \map \sin {\paren {0 + 1} \theta}
      | r = \map \sin \theta
}}
{{eqn | r = 1 \cdot \sin \theta
}}
{{end-eqn}}
:$\map {U_0} x = 1$, $U_0 \in \Bbb P$
{{MissingLinks|$\Bbb P$ as the set/space of polynomials}}
For $n = 1$:
{{begin-eqn}}
{{eqn | l = \map \sin {\paren {1 + 1} \theta}
      | r = \map \sin {2 \theta}
}}
{{eqn | r = 2 \sin \theta \cos \theta
      | c = Double Angle Formula for Sine
}}
{{eqn | r = 2 \cos \theta \cdot \sin \theta
}}
{{end-eqn}}
:$\map {U_1} x = 2 x$, $U_1 \in \Bbb P$
Assume $\map {U_n} {\cos \theta} \sin \theta = \map \sin {\paren {n + 1} \theta}$.
{{begin-eqn}}
{{eqn | l = \map {U_{n + 1} } {\cos \theta} \sin \theta
      | r = \map \sin {\paren {n + 1 + 1} \theta}
}}
{{eqn | r = \map \sin {\paren {n + 1} \theta + \theta}
}}
{{eqn | r = \map \sin {\paren {n + 1} \theta} \map \cos \theta + \map \cos {\paren {n + 1} \theta} \map \sin \theta
      | c = Sine of Sum
}}
{{eqn | r = 2 \map \sin {\paren {n + 1} \theta} \map \cos \theta + \map \cos {\paren {n + 1} \theta} \map \sin \theta - \map \sin {\paren {n + 1} \theta} \map \cos \theta
}}
{{eqn | r = 2 \map \sin {\paren {n + 1} \theta} \map \cos \theta - \paren {\map \sin {\paren {n + 1} \theta} \map \cos \theta - \map \cos { \paren {n + 1} \theta} \map \sin \theta}
}}
{{eqn | r = 2 \map \sin {\paren {n + 1} \theta} \map \cos \theta - \map \sin {\paren {n + 1} \theta - \theta}
      | c = Sine of Difference
}}
{{eqn | r = 2 \map \cos \theta \map \sin {\paren {n + 1} \theta} - \map \sin {\paren {n - 1 + 1} \theta}
}}
{{eqn | r = 2 \map \cos \theta \map {U_n} {\cos \theta} \map \sin \theta - \map {U_{n - 1} } {\cos \theta} \sin \theta
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {2 \map \cos \theta \map {U_n} {\cos \theta} - \map {U_{n - 1} } {\cos \theta} } \cdot \map \sin \theta
}}
{{end-eqn}}
:$\map {U_{n + 1} } x = 2 x \map {U_n} x - \map {U_{n - 1} } x$, $U_{n + 1} \in \Bbb P$
By the Second Principle of Mathematical Induction, $U_n \in \Bbb P$ for all natural numbers $n$.
{{qed}}
Category:Chebyshev Polynomials of the Second Kind
hv955hz5901ppfwhb2ndepcbtki3z2p
Theorem: Composition of Relations Preserves Subsets

Let $A, B, S, T$ be relations as subsets of Cartesian products.
Let $A \subseteq B$ and $S \subset T$.
Then:
:$A \circ S \subseteq B \circ T$

Proof:
We have:
{{begin-eqn}}
{{eqn | q = \forall \tuple{x, y} 
      | l = \tuple{x, y} \in A \circ S
      | o = \leadsto
      | r = \exists z : \tuple{x, z} \in S, \tuple{z, y} \in A
      | c = {{Defof|Composite Relation}}
}}
{{eqn | o = \leadsto
      | r = \exists z : \tuple{x, z} \in T, \tuple{z, y} \in B
      | c = {{Defof|Subset}}
}}
{{eqn | o = \leadsto
      | r = \tuple{x, y} \in B \circ T
      | c = {{Defof|Composite Relation}}
}}
{{end-eqn}}
By definition of subset:
:$A \circ S \subseteq B \circ T$
{{qed}}
Category:Relations
Category:Cartesian Product
Category:Composite Relations
qomu5rxk5z0vj6zn0luxa4drwpy82pu
Theorem: Primitive of Reciprocal of Root of a x + b by Root of p x + q/a greater than 0, p less than 0

Let $a > 0$ and $p < 0$.
Then:
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C$
for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.

Proof:
=== Completing the Square ===
{{:Primitive of Reciprocal of Root of a x + b by Root of p x + q/Completing the Square}}{{qed|lemma}}
Hence:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
      | r = \frac 1 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {b p - a q}^2 - u^2} }
      | c = Completing the Square above, as $a p < 0$
}}
{{eqn | r = \frac 1 {\sqrt {-a p} } \map \arcsin {\dfrac u {b p - a q} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$
}}
{{eqn | r = \frac 1 {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {b p - a q} } + C
      | c = substituting for $u$ from Completing the Square above
}}
{{eqn-intertext|We could finish here, but we wish to present the result in the format given, so:}}
{{eqn | r = \frac {-1} {\sqrt {-a p} } \map \arcsin {-\dfrac {2 a p x + b p + a q} {b p - a q} } + C
      | c = Arcsine is Odd Function
}}
{{eqn | r = \frac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C
      | c = simplifying
}}
{{end-eqn}}
</onlyinclude>{{qed}}
Theorem: Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p greater than 0

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 $1$ ===
{{:Primitive of Reciprocal of Root of a x + b by Root of p x + q/Lemma 1}}{{qed|lemma}}
Theorem: Characterization of Paracompactness in T3 Space/Lemma 15

:$\forall n \in \N_{>0}: \set{\map {U_n} x : x \in X}$ refines $\set{\map {V_0} x : x \in X}$

Proof:
From Composite of Reflexive Relations is Reflexive:
:$\forall n \in \N_{>0} : U_n$ is reflexive.
From Set of Images of Reflexive Relation is Cover of Set:
:$\set{\map {V_0} x : x \in X}$ is a cover of $X$.
and
:$\forall n \in \N_{>0} : \set{\map {U_n} x : x \in X}$ is a cover of $X$.
Theorem: Set of Images of Reflexive Relation is Cover of Set

Let $\RR \subseteq S \times S$ be a reflexive relation in $S$.
Let $\II = \set{\map \RR x : x \in S}$ be the set of images under $\RR$. 
Then:
:$\II$ is a cover of $S$

Proof:
By definition of reflexive relation:
:$\forall x \in S : \tuple{x, x} \in \RR$
By definition of image:
:$\forall x \in S : x \in \map \RR x$
Hence, $\II$ is a cover of $S$ by definition.
{{qed}}
Category:Reflexive Relations
Category:Covers
pkki8rprjko873pvq5fy5roibdawvle
Theorem: Image of Point under Neighborhood of Diagonal is Neighborhood of Point

Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $V$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall x \in X : \map V x$ is a neighborhood of $x$ in $T$

Proof:
Let $x \in X$.
By definition of diagonal:
:$\tuple{x, x} \in \Delta_X$
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$V$ is a neighborhood of $\tuple{x, x}$
By definition of the product topology:
:$\BB = \set {U_1 \times U_2: U_1, U_2 \in \tau}$ is a basis for $\tau_{X \times X}$
From Characterization of Neighborhood by Basis:
:$\exists U_1, U_2 \in \tau : \tuple{x, x} \in U_1 \times U_2$ and $U_1 \times U_2 \subseteq V$
By definition of Cartesian product:
:$x \in U_1, x \in U_2$
We have:
{{begin-eqn}}
{{eqn | l = U_2
      | r = \map {\paren{U_1 \times U_2} } x
      | c = Image of Element under Cartesian Product of Subsets
}}
{{eqn| o = \subseteq
      | r = \map V x
      | c = {{Corollary|Image under Subset of Relation is Subset of Image under Relation}}
}}
{{end-eqn}}
Hence $\map V x$ is a neighborhood of $x$ in $T$ by definition.
The result follows.
{{qed}}
Category:Neighborhoods
Category:Product Spaces
Category:Relations
5ox734mp81ndc1gx1fhmm4myo58lyp2
Theorem: Image of Element under Cartesian Product of Subsets

Let $S$ and $T$ be sets.
Let $A \subseteq S$ and $B \subseteq T$.
Let $\RR$ be the relation defined by the Cartesian product $A \times B$.
Then:
:$\forall x \in A: \map \RR x = B$

Proof:
We have:
{{begin-eqn}}
{{eqn | q = \forall x \in A
      | l = \map \RR s
      | r = \set {t \in T: \tuple {s, t} \in \RR}
      | c = {{Defof|Image of Element under Relation}}
}}
{{eqn | r = \set {t \in T: \tuple {s, t} \in A \times B}
      | c = Definition of $\RR$
}}
{{eqn | r = \set {t \in T: s \in A, t \in B}
      | c = {{Defof|Cartesian Product}}
}}
{{eqn | r = \set {t \in T: t \in B}
      | c = as $s \in A$
}}
{{eqn | r = B
      | c = as $B \subseteq T$
}}
{{end-eqn}}
{{qed}}
Category:Cartesian Product
Category:Relations
dond7fj1vd0xt7riklj1g70jech9byx
Theorem: Image under Subset of Relation is Subset of Image under Relation/Corollary

Let $x \in S$.
Then:
:$\map {\RR_2} x \subseteq \map {\RR_1} x$
where $\map {\RR_1} x$ denotes the image of $x$ under $\RR_1$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\RR_2} x
      | r = \RR_2 \sqbrk {\set x}
      | c = Image of Singleton under Relation
}}
{{eqn | o = \subseteq 
      | r = \RR_1 \sqbrk {\set x}
      | c = Image under Subset of Relation is Subset of Image under Relation
}}
{{eqn | r = \map {\RR_1} x
      | c = Image of Singleton under Relation
}}
{{end-eqn}}
{{qed}}
Category:Relation Theory
1r5rzsid6uq3notj230d4y1jcufay54
Theorem: Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset

Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $V$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall A \subseteq X : V \sqbrk A$ is a neighborhood of $A$ in $T$

Proof:
Let $A \subseteq X$.
From Image of Subset under Relation equals Union of Images of Elements:
:$V \sqbrk A = \ds \bigcup_{x \in A} \map V x$
From Subset of Union:
:$\forall x \in A : \map V x \subseteq V \sqbrk A$
From Image of Point under Neighborhood of Diagonal is Neighborhood of Point:
:$\forall x \in A : \map V x$ is a neighborhood of $x$ in $T$
From Superset of Neighborhood in Topological Space is Neighborhood:
:$\forall x \in A : \map V A$ is a neighborhood of $x$ in $T$
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$\map V A$ is a neighborhood of $A$ in $T$
{{qed}}
Category:Neighborhoods
Category:Product Spaces
Category:Relations
pz0p9z254w2ydnywje8f5pra0jvvua5
Theorem: Lemmata for Euler's Third Substitution/Lemma 1

:$x = \dfrac {a \beta - \alpha t^2} {a - t^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt {a x^2 + b x + c}
      | r = \sqrt {a \paren {x - \alpha} \paren {x - \beta} }
      | c = 
}}
{{eqn | r = \paren {x - \alpha} t
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = a \paren {x - \alpha} \paren {x - \beta}
      | r = \paren {x - \alpha}^2 t^2
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = a \paren {x - \beta}
      | r = \paren {x - \alpha} t^2
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = a x - a \beta
      | r = x t^2 - \alpha t^2
      | c = multiplying out
}}
{{eqn | ll= \leadsto
      | l = x \paren {a - t^2}
      | r = a \beta - \alpha t^2
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \dfrac {a \beta - \alpha t^2} {a - t^2}
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Lemmata for Euler's Third Substitution/Lemma 2

:$x - \alpha = \dfrac {a \paren {\alpha - \beta} } {t^2 - a}$

Proof:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac {a \beta - \alpha t^2} {a - t^2}
      | c = {{Lemma|Lemmata for Euler's Third Substitution|1|proof = yes}}
}}
{{eqn | ll= \leadsto
      | l = x - \alpha
      | r = \dfrac {a \beta - \alpha t^2} {a - t^2} - \alpha
      | c = 
}}
{{eqn | r = \dfrac {a \beta - \alpha t^2 - \alpha \paren {a - t^2} } {a - t^2}
      | c = common denominator
}}
{{eqn | r = \dfrac {a \paren {\alpha - \beta} } {t^2 - a}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Lemmata for Euler's Third Substitution/Lemma 4

:$t = \pm \sqrt {\dfrac {a \paren {x - \beta} } {x - \alpha} }$

Proof:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac {a \beta - \alpha t^2} {a - t^2}
      | c = {{Lemma|Lemmata for Euler's Third Substitution|1|proof = yes}}
}}
{{eqn | ll= \leadsto
      | l = x \paren {a - t^2}
      | r = a \beta - \alpha t^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = t^2 \paren {x - \alpha}
      | r = a \paren {x - \beta}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = t^2
      | r = \dfrac {a \paren {x - \beta} } {x - \alpha}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = t
      | r = \pm \sqrt {\dfrac {a \paren {x - \beta} } {x - \alpha} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Lemmata for Euler's Third Substitution/Lemma 3

:$\dfrac {\d x} {\d t} = \dfrac {2 t a \paren {\beta - \alpha} } {\paren {a - t^2}^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {\d x} {\d t}
      | r = \map {\dfrac \d {\d t} } {x - \alpha}
      | c = Derivative of Function plus Constant
}}
{{eqn | r = \map {\dfrac \d {\d t} } {\dfrac {a \paren {\alpha - \beta} } {t^2 - a} }
      | c = {{Lemma|Lemmata for Euler's Third Substitution|2|proof = yes}}
}}
{{eqn | r = \dfrac {-a \paren {\alpha - \beta} } {\paren {t^2 - a}^2} \cdot \paren {2 t}
      | c = Power Rule for Derivatives, Chain Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d x} {\d t}
      | r = \dfrac {2 t a \paren {\beta - \alpha} } {\paren {a - t^2}^2}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) greater than 0

Let $p \paren {b p - a q} > 0$.
Then:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \frac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$

Proof:
=== Lemma ===
{{:Primitive of Reciprocal of p x + q by Root of a x + b/Lemma}}{{qed|lemma}}
We have {{hypothesis}} that:
:$p \paren {b p - a q} > 0$
which means:
:$\dfrac {b p - a q} p > 0$
Hence let:
:$d^2 = \dfrac {b p - a q} p$
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }
      | r = \frac 2 p \int \frac {\d u} {u^2 - \paren {\dfrac {b p - a q} p} }
      | c = {{Lemma|Primitive of Reciprocal of p x + q by Root of a x + b}}
}}
{{eqn | r = \frac 2 p \int \frac {\d u} {u^2 - d^2}
      | c = setting $d^2 = \dfrac {b p - a q} p$
}}
{{eqn | r = \frac 2 p \frac 1 {2 d} \ln \size {\frac {u - d} {u + d} } + C
      | c = Primitive of $\dfrac 1 {u^2 - d^2}$: Logarithm Form
}}
{{eqn | r = \frac 2 p \dfrac 1 2 \sqrt {\dfrac p {b p - a q} } \ln \size {\frac {u - \sqrt {\dfrac {b p - a q} p } } {u + \sqrt {\dfrac {b p - a q} p} } } + C
      | c = substituting for $d$
}}
{{eqn | r = \frac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\frac {\sqrt {p u} - \sqrt {b p - a q} } {\sqrt {p u} + \sqrt {b p - a q} } } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 {\sqrt {p \paren {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
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) less than 0

Let $p \paren {b p - a q} > 0$.
Then:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C$

Proof:
=== Lemma ===
{{:Primitive of Reciprocal of p x + q by Root of a x + b/Lemma}}{{qed|lemma}}
We have {{hypothesis}} that:
:$p \paren {b p - a q} < 0$
which means:
:$\dfrac {b p - a q} p < 0$
Hence let:
:$d^2 = -\dfrac {b p - a q} p$
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }
      | r = \frac 2 p \int \frac {\d u} {u^2 + \paren {-\dfrac {b p - a q} p} }
      | c = {{Lemma|Primitive of Reciprocal of p x + q by Root of a x + b}}
}}
{{eqn | r = \frac 2 p \int \frac {\d u} {u^2 + d^2}
      | c = setting $d^2 = \dfrac {b p - a q} p$
}}
{{eqn | r = \frac 2 p \frac 1 d \arctan {\frac u d} + C
      | c = Primitive of $\dfrac 1 {u^2 + d^2}$
}}
{{eqn | r = \frac 2 p \sqrt {-\dfrac p {b p - a q} } \arctan {\sqrt {-\dfrac p {b p - a q} } \sqrt {a x + b} } + C
      | c = substituting for $d$ and $u$
}}
{{eqn | r = \dfrac {-2} {\sqrt {p \paren {a q - b p} } } \arctan {-\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C
      | c = simplifying
}}
{{eqn | r = \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan {\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C
      | c = Arctangent Function is Odd
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Root of a x squared plus b x plus c/Lemma

:$\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } = -\int \frac {\d u} {\pm \sqrt {a + b u + c u^2} }$
where $u := \dfrac 1 x$, according to whether $u > 0$ or $u < 0$.

Proof:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac 1 u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = \frac {-1} {u^2}
      | c = Primitive of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | r = \int {\frac 1 {\frac 1 u \sqrt {a \paren {\frac 1 u}^2 + b \paren {\frac 1 u} + c} } } \frac {-\rd u} {u^2}
      | c = Integration by Substitution
}}
{{eqn | r = -\int \frac {\d u} {\frac {u^2} u \sqrt {\frac 1 {u^2} \paren {a + b u + c u^2} } }
      | c = Primitive of Constant Multiple of Function and simplification
}}
{{eqn | r = -\int \frac {\d u} {\frac {u^2} u \sqrt {\frac 1 {u^2} } \sqrt {a + b u + c u^2} }
      | c = extracting $\sqrt {\frac 1 {u^2} }$
}}
{{eqn | r = -\int \frac {\d u} {\pm \sqrt {a + b u + c u^2} }
      | c = according to whether $u > 0$ or $u < 0$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Root of x squared minus a squared/Arccosine Form

:$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arccos \size {\frac a x} + C$
for $0 < a < \size x$.

Proof:
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
:$x > a$
or:
:$x < -a$
where it is assumed that $a > 0$.
Hence:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \sqrt {x^2 - a^2} }
      | r = \frac 1 a \arcsec \size {\frac x a} + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$: Arcsecant Form
}}
{{eqn | r = \frac 1 a \arccos \size {\frac a x} + C
      | c = Arcsecant of Reciprocal equals Arccosine
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Root of 2 a x minus x squared

:$\ds \int \frac {\d x} {\sqrt {2 a x - x^2} } = \arcsin \dfrac {x - a} a + C$

Proof:
Let $u := x - a$.
Then:
:$\dfrac {\d u} {\d x} = 1$
and:
:$x = u + a$
and we have:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\sqrt {2 a x - x^2} }
      | r = \int \frac {\d u} {\sqrt {2 a \paren {u + a} - \paren {u + a}^2} }
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac {\d u} {\sqrt {2 a u + 2 a^2 - u^2 - 2 a u - a^2} }
      | c = multiplying out
}}
{{eqn | r = \int \frac {\d u} {\sqrt {a^2 - u^2} }
      | c = simplifying
}}
{{eqn | r = \arcsin \frac u a + C
      | c = Primitive of $\dfrac 1 {\sqrt {a^2 - u^2} }$: Arcsine Form
}}
{{eqn | r = \arcsin \dfrac {x - a} a + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of 2 a x minus x squared

:$\ds \int \sqrt {2 a x - x^2} \rd x = \frac {\paren {x - a} } 2 \sqrt {2 a x - x^2} + \frac {a^2} 2 \arcsin \frac {x - a} a + C$

Proof:
Let $u := x - a$.
Then:
:$\dfrac {\d u} {\d x} = 1$
and:
:$x = u + a$
Then:
{{begin-eqn}}
{{eqn | l = 2 a x - x^2
      | r = 2 a \paren {u + a} - \paren {u + a}^2
      | c = 
}}
{{eqn | r = 2 a u + 2 a^2 - u^2 - 2 a u - a^2
      | c = 
}}
{{eqn | r = a^2 - u^2
      | c = 
}}
{{end-eqn}}
and we have:
{{begin-eqn}}
{{eqn | l = \int \sqrt {2 a x - x^2} \rd x
      | r = \int \sqrt {a^2 - u^2} \rd u
      | c = 
}}
{{eqn | r = \frac {u \sqrt {a^2 - u^2} } 2 + \frac {a^2} 2 \arcsin \frac u a + C
      | c = Primitive of $\sqrt {a^2 - u^2}$: Arcsine Form
}}
{{eqn | r = \frac {\paren {x - a} } 2 \sqrt {2 a x - x^2} + \frac {a^2} 2 \arcsin \frac {x - a} a + C
      | c = substituting for $u$ and simplifying
}}
{{end-eqn}}
{{qed}}