avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of a x + b over p x + q

:$\ds \int \frac {a x + b} {p x + q} \rd x  = \frac {a x} p + \frac {b p - a q} {p^2} \ln \size {p x + q} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {a x + b} {p x + q} \rd x
      | r = a \int \frac {x \rd x} {p x + q} + b \int \frac {\d x} {p x + q}
      | c = Linear Combination of Primitives
}}
{{eqn | r = a \paren {\frac x p - \frac q {p^2} \ln \size {p x + q} } + b \int \frac {\d x} {p x + q} + C
      | c = Primitive of $\dfrac x {a x + b}$
}}
{{eqn | r = \frac {a x} p - \frac {a q} {p^2} \ln \size {p x + q} + b \paren {\frac 1 p \ln \size {p x + q} } + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | r = \frac {a x} p - \frac {a q} {p^2} \ln \size {p x + q} + \frac b p \ln \size {p x + q} + C
      | c = 
}}
{{eqn | r = \frac {a x} p + \frac {b p - a q} {p^2} \ln \size {p x + q} + C
      | c = common denominator
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of a x + b over Power of p x + q/Formulation 1

:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac {\paren {a x + b}^{m + 1} } {\paren {p x + q}^{n - 1} } + \paren {n - m - 2} a \int \frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } \rd x}$

Proof:
From Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - \paren {m + n + 2} a \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$
Setting $n := -n$:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x
      | c = 
}}
{{eqn | r = \int \paren {a x + b}^m \paren {p x + q}^{-n} \rd x
      | c = 
}}
{{eqn | r = \frac 1 {\paren {-n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{-n + 1} - \paren {m + -n + 2} a \int \paren {a x + b}^m \paren {p x + q}^{-n + 1} \rd x}
      | c = 
}}
{{eqn | r = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac {\paren {a x + b}^{m + 1} } {\paren {p x + q}^{n - 1} } + \paren {n - m - 2} a \int \frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of a x + b over Power of p x + q/Formulation 2

:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$

Proof:
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} a} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$
Setting $n := -n$:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x
      | c = 
}}
{{eqn | r = \int \paren {a x + b}^m \paren {p x + q}^{-n} \rd x
      | c = 
}}
{{eqn | r = \frac {\paren {a x + b}^m \paren {p x + q}^{-n + 1} } {\paren {m - n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m - n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^{-n} \rd x
      | c = 
}}
{{eqn | r = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m  \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of a x + b over Power of p x + q/Formulation 3

:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^{n - 1}} \rd x}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \paren {a x + b}^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = m a \paren {a x + b}^m
      | c = Power Rule for Derivatives and Derivatives of Function of $a x + b$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = v
      | r = \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = \paren {p x + q}^n
      | c = Power Rule for Derivatives and Derivatives of Function of $a x + b$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x
      | c = 
}}
{{eqn | r = \paren {a x + b}^m \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } - \int \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } m a \paren {a x + b}^m \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m-1} } {\paren {p x + q}^{n - 1} } \rd x}
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of p x + q over Root of a x + b

:$\ds \int \frac {p x + q} {\sqrt {a x + b} } \rd x = \frac {2 \paren {a p x + 3 a q - 2 b p} } {3 a^2} \sqrt{a x + b}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \sqrt{a x + b}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {u^2 - b} a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {p x + q} {\sqrt {a x + b} } \rd x
      | r = \int \paren {p \paren {\frac {u^2 - b} a} + q} \frac {2 u} {a u} \rd x
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 {a^2} \int \paren {p u^2 - b p + a q} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 2 {a^2} \paren {\frac {p u^3} 3 - b p z + a q z} + C
      | c = Primitive of Power
}}
{{eqn-intertext|Note that at this point we can assume $u > 0$, otherwise $\sqrt {a x + b}$ would not have been defined.}}
{{eqn | r = \frac 2 {a^2} \paren {\frac {p \paren {\sqrt {a x + b} }^3} 3 - b p \sqrt {a x + b} + a q \sqrt {a x + b} } + C
      | c = substituting for $u$
}}
{{eqn | r = \frac {2 \paren {a p x - 2 b p + 3 a q} } {3 a^2} \sqrt {a x + b} + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p x + q by Root of a x + b

Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.
Then:
:<nowiki>$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases}
\dfrac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C & : p \paren {b p - a q} > 0 \\ \\
\dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C & : p \paren {b p - a q} < 0 \\
\end {cases}$</nowiki>

Proof:
=== Case $1$: $p \paren {b p - a q} > 0$ ===
{{:Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) greater than 0}}{{qed|lemma}}
Theorem: Primitive of Root of a x + b over p x + q

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

Proof:
From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 2:
:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$
Setting $m := \dfrac 1 2$ and $n = 1$:
{{begin-eqn}}
{{eqn | l = \int \frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^n} \rd x
      | r = \frac {-1} {\paren {1 - \frac 1 2 - 1} p} \paren {\frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^0} + \frac 1 2 \paren {b p - a q} \int \frac {\paren {a x + b}^{1/2 - 1} } {\paren {p x + q} } \rd x}
      | c = 
}}
{{eqn | r = \frac {2 \sqrt {a x + b} } p + \frac {b p - a q} p \int \frac {\d x} {\sqrt {a x + b} \paren {p x + q} }
      | c = 
}}
{{end-eqn}}
From Primitive of Reciprocal of $p x + q$ by Root of $a x + b$:
:<nowiki>$\ds \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} } = \begin{cases}
\dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \\
\dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\
\end{cases}$</nowiki>
The result follows by substitution.
{{qed}}
Theorem: Primitive of Power of p x + q by Root of a x + b

:$\ds \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt{a x + b} } \rd x$

Proof:
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$
Setting $m := \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \paren {a x + b}^{1 / 2} \paren {p x + q}^n \rd x
      | r = \frac {\paren {a x + b}^{1 / 2} \paren {p x + q}^{n + 1} } {\paren {\frac 1 2 + n + 1} p} + \frac {\dfrac 1 2 \paren {b p - a q} } {\paren {\frac 1 2 + n + 1} p} \int \paren {a x + b}^{1 / 2 - 1} \paren {p x + q}^n \rd x
      | c = 
}}
{{eqn | r = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of p x + q by Root of a x + b

:$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$

Proof:
From Primitive of Reciprocal of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$:
:$\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$
Setting $m := \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} }
      | r = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{1/2 - 1} \paren {p x + q}^{n - 1} } + a \paren {\frac 1 2 + n - 2} \int \frac {\d x} {\paren {a x + b}^{1/2} \paren {p x + q}^{n - 1} } }
      | c = 
}}
{{eqn | r = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of p x + q over Root of a x + b

:$\ds \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$

Proof:
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$
Setting $m := -\dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x
      | r = \frac {\paren {a x + b}^{-1/2 + 1} \paren {p x + q}^n} {\paren {-\frac 1 2 + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {-\frac 1 2 + n + 1} a} \int \paren {a x + b}^{-1/2} \paren {p x + q}^{n - 1} \rd x
      | c = 
}}
{{eqn | r = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a x + b over Power of p x + q

:$\ds \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x = \frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$

Proof:
From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 3:
:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m a \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^{n - 1} } \rd x}$
Setting $m := \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x
      | r = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^{n - 1} } + \frac 1 2 a \int \frac {\paren {a x + b}^{1/2 - 1} } {\paren {p x + q}^{n - 1} } \rd x}
      | c = 
}}
{{eqn | r =\frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Root of a x + b by Root of p x + q

Let $a, b, p, q \in \R$ such that $a p \ne b q$.
Then:
:<nowiki>$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C & : a p < 0 \\ \\
\dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C & : a > 0, p < 0 \\ \\
\ds \dfrac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C & : a p > 0
\end {cases}$</nowiki>
for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.

Proof:
=== Case $1$: $a p < 0$ ===
{{:Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0}}{{qed|lemma}}
Theorem: Derivative of Laplace Transform

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any closed interval $\closedint 0 a$.
Let $\laptrans f = F$ denote the Laplace transform of $f$.
Then, everywhere that $\dfrac \d {\d s} \laptrans f$ exists:
:$\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$

Proof:
{{mistake|see talk page}}
{{begin-eqn}}
{{eqn | l = \frac \d {\d s} \laptrans {\map f t}
      | r = \frac \d {\d s} \int_0^{\to +\infty} \map f t \, e^{-s t} \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \int_0^{\to +\infty} \map {\frac {\partial} {\partial s} } {\map f t \, e^{-s t} } \rd t
      | c = Definite Integral of Partial Derivative
}}
{{eqn | r = \int_0^{\to +\infty} \map f t \, \map {\frac {\partial} {\partial s} } {e^{-st} } \rd t
}}
{{eqn | r = -\int_0^{\to +\infty} t \, \map f t \, e^{-s t} \rd t
      | c = Derivative of Exponential Function
}}
{{eqn | r = -\laptrans {t \, \map f t}
      | c = {{Defof|Laplace Transform}}
}}
{{end-eqn}}
{{qed}}
Theorem: Higher Order Derivatives of Laplace Transform

:$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$

Proof:
The proof proceeds by induction on $n$, the order of the derivative of $\laptrans f$.
The case $n = 0$ is verified as follows:
{{begin-eqn}}
{{eqn | l = \frac {\d^0} {\d s^0} \laptrans {\map f t}
      | r = \laptrans {\map f t}
      | c = {{Defof|Zeroth Derivative}}
}}
{{eqn | r = \paren {-1}^0 \laptrans {t^0 \, \map f t}
      | c = {{Defof|Power of Natural Number|Zeroth Power}}
}}
{{end-eqn}}
Theorem: Primitive of Root of a x + b by Root of p x + q

:$\ds \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$

Proof:
From Primitive of $\paren {p x + q}^n \sqrt {a x + b}$:
:$\ds \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x$
Putting $n = \dfrac 1 2$:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x
      | c = 
}}
{{eqn | r = \frac {2 \paren {p x + q}^{1 / 2 + 1} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 3} p} + \frac {b p - a q} {\paren {2 \cdot \frac 1 2 + 3} p} \int \frac {\paren {p x + q}^{1 / 2} } {\sqrt {a x + b} } \rd x
      | c = 
}}
{{eqn | r = \frac {2 \paren {p x + q} \sqrt {p x + q} \sqrt {a x + b} } {4 p} + \frac {b p - a q} {4 p} \int \frac {\sqrt {p x + q} } {\sqrt {a x + b} } \rd x
      | c = 
}}
{{eqn | r = \frac {2 \paren {p x + q} \sqrt {p x + q} \sqrt {a x + b} } {4 p} + \frac {b p - a q} {4 p} \paren {\frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } }
      | c = Primitive of $\dfrac {\sqrt {p x + q} } {\sqrt {a x + b} }$
}}
{{eqn | r = \frac {\paren {2 a \paren {p x + q} + \paren {b p - a q} } \sqrt {\paren {a x + b} \paren {p x + q} } } {4 a p} - \frac {\paren {b p - a q}^2 } {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
      | c = extracting common factor
}}
{{eqn | r = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of p x + q over Root of a x + b

:$\ds \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$

Proof:
From Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$:
:$\ds \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$
Putting $n = \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \sqrt {\frac {p x + q} {a x + b} } \rd x
      | r = \frac {2 \paren {p x + q}^{1/2} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 1} a} + \frac {2 \cdot \frac 1 2 \paren {a q - b p} } {\paren {2 \cdot \frac 1 2 + 1} a} \int \frac {\paren {p x + q}^{1/2 - 1} } {\sqrt {a x + b} } \rd x
      | c = 
}}
{{eqn | r = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p x + q by Root of a x + b by Root of p x + q

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

Proof:
From Primitive of $\paren {p x + q}^n \sqrt {a x + b}$:
:$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} }$
Putting $n = \dfrac 3 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }
      | r = \frac {\sqrt {a x + b} } {\paren {\frac 3 2 - 1} \paren {a q - b p} \paren {p x + q}^{3 / 2 - 1} } + \frac {\paren {2 \cdot \frac 3 2 - 3} a} {2 \paren {\frac 3 2 - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{3 / 2 - 1} \sqrt {a x + b} }
      | c = 
}}
{{eqn | r = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + \frac {\paren {3 - 3} a} {a q - b p} \int \frac {\d x} {\sqrt {p x + q} \sqrt {a x + b} } + C
      | c = 
}}
{{eqn | r = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x squared plus a squared

:$\ds \int \frac {x \rd x} {x^2 + a^2} = \frac 1 2 \map \ln {x^2 + a^2} + C$

Proof:
{{begin-eqn}}
{{eqn | l = u
      | r = x^2 + a^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 2 x 
      | c = Power Rule for Derivatives and Derivative of Constant
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {x^2 + a^2}
      | r = \frac 1 2 \ln \size {x^2 + a^2} + C
      | c = Primitive of Function under its Derivative
}}
{{eqn | r = \frac 1 2 \, \map \ln {x^2 + a^2} + C
      | c = Absolute Value of Even Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x squared plus a squared

:$\ds \int \frac {x \rd x} {x^2 + a^2} = \frac 1 2 \map \ln {x^2 + a^2} + C$

Proof:
From Primitive of Power of x less one over Power of x plus Power of a:
:$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$
So:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {x^2 + a^2}
      | r = \frac 1 2 \ln \size {x^2 + a^2} + C
      | c = Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 2$
}}
{{eqn | r = \frac 1 2 \, \map \ln {x^2 + a^2} + C
      | c = Absolute Value of Even Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x squared plus a squared

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {x^2 + a^2}
      | r = \int \paren {1 - \frac {a^2} {x^2 + a^2} } \rd x
      | c = long division
}}
{{eqn | r = \int \d x - a^2 \int \frac {\d x} {x^2 + a^2} 
      | c = Linear Combination of Primitives
}}
{{eqn | r = x - a^2 \int \frac {\d x} {x^2 + a^2} + C
      | c = Primitive of Constant
}}
{{eqn | r = x - a^2 \frac 1 a \arctan {\frac x a} + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = x - a \arctan {\frac x a} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x squared plus a squared

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = a \tan \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \sec^2 \theta
      | c = Derivative of Tangent Function
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}
      | r = \int \frac {a^2 \tan^2 \theta a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac {a^3 \tan^2 \theta \sec^2 \theta} {a^4 \sec^4 \theta} \rd \theta
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \frac 1 a \int \frac {\tan^2 \theta} {\sec^2 \theta} \rd \theta
      | c = simplification
}}
{{eqn | r = \frac 1 a \int \tan^2 \theta \cos^2 \theta \rd \theta
      | c = {{Defof|Real Secant Function}}
}}
{{eqn | r = \frac 1 a \int \frac {\sin^2 \theta} {\cos^2 \theta} \cos^2 \theta \rd \theta
      | c = {{Defof|Real Tangent Function}}
}}
{{eqn | r = \frac 1 a \int \sin^2 \theta \rd \theta
      | c = simplification
}}
{{eqn | r = \frac \theta {2 a} - \frac {\sin 2 \theta} {4 a} + C
      | c = Primitive of Square of Sine Function
}}
{{eqn | r = \frac \theta {2 a}  - \frac 1 {2 a} \frac {\tan \theta} {1 + \tan^2 \theta} + C
      | c = Tangent Half-Angle Substitution for Sine
}}
{{eqn | r = -\frac x {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C
      | c = substituting for $\theta$
}}
{{end-eqn}}
{{qed}}