avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Root of a x + b over Power of x/Formulation 2

:$\ds \int \frac {\sqrt{a x + b} } {x^m} \rd x = -\frac {\paren {\sqrt{a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 5} a} {\paren {2 m - 2} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x$

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Putting $n := \dfrac 1 2$ and $m := -m$:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } {x^m} \rd x
      | r = \int x^{-m} \paren {a x + b}^{1 / 2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^{-m + 1} \paren {a x + b}^{1 / 2 + 1} } {\paren {-m + 1} b} - \frac {\paren {-m + \frac 1 2 + 2} a} {\paren {-m + 1} b} \int x^{-m + 1} \paren {a x + b}^{1 / 2} \rd x
      | c = 
}}
{{eqn | r = -\frac {\paren {\sqrt {a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {m - \frac 5 2} a} {\paren {m - 1} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x
      | c = simplifying
}}
{{eqn | r = -\frac {\paren {\sqrt {a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 5} a} {\paren {2 m - 2} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of Root of a x + b

:$\ds \int \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 2} } {a \paren {m + 2} } + C$
for $m \ne -2$.

Proof:
Let $u = \sqrt {a x + b}$.
Then:
{{begin-eqn}}
{{eqn | l = \int \paren {\sqrt {a x + b} }^m \rd x
      | r = \frac 2 a \int u \cdot u^m \rd x
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 a \int u^{m + 1} \rd x
      | c = simplifying
}}
{{eqn | r = \frac 2 a \frac {u^{m + 2} } {m + 2} + C
      | c = Primitive of Power valid for $m + 2 \ne 0$, that is, $m \ne -2$
}}
{{eqn | r = \frac {2 \paren {\sqrt {a x + b} }^{m + 2} } {a \paren {m + 2} } + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Power of Root of a x + b

:$\ds \int x \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 4} } {a^2 \paren {m + 4} } - \frac {2 b \paren {\sqrt {a x + b} }^{m + 2} } {a^2 \paren {m + 2} } + C$

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 x \paren {\sqrt {a x + b} }^m \rd x
      | r = \frac 2 a \int \frac {u^2 - b} a u^{m + 1} \rd x
      | c = Primitive of Function of $a x + b$
}}
{{eqn | r = \frac 2 {a^2} \int u^{m + 3} \rd x - \frac {2 b} {a^2} \int u^{m + 1} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 2 {a^2} \frac {u^{m + 4} } {m + 4} - \frac {2 b} {a^2} \frac {u^{m + 2} } {m + 2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {2 \paren {\sqrt {a x + b} }^{m + 4} } {a^2 \paren {m + 4} } - \frac {2 b \paren {\sqrt {a x + b} }^{m + 2} } {a^2 \paren {m + 2} } + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Power of Root of a x + b

:$\ds \int x^2 \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 6} } {a^3 \paren {m + 6} } - \frac {4 b \paren {\sqrt {a x + b} }^{m + 4} } {a^3 \paren {m + 4} } + \frac {2 b^2 \paren {\sqrt {a x + b} }^{m + 2} } {a^3 \paren {m + 2} } + C$

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 x^2 \paren {\sqrt {a x + b} }^m \rd x
      | r = \frac 2 a \int \paren {\frac {u^2 - b} a}^2 u^{m + 1} \rd x
      | c = Primitive of Function of $a x + b$
}}
{{eqn | r = \frac 2 {a^3} \int u^{m + 5} \rd x - \frac {4 b} {a^3} \int u^{m + 3} \rd x + \frac {2 b^2} {a^3} \int u^{m + 1} \rd x 
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 2 {a^3} \frac {u^{m + 6} } {m + 6} - \frac {4 b} {a^3} \frac {u^{m + 4} } {m + 4} + \frac {2 b^2} {a^3} \frac {u^{m + 2} } {m + 2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {2 \paren {\sqrt {a x + b} }^{m + 6} } {a^3 \paren {m + 6} } - \frac {4 b \paren {\sqrt {a x + b} }^{m + 4} } {a^3 \paren {m + 4} } + \frac {2 b^2 \paren {\sqrt {a x + b} }^{m + 2} } {a^3 \paren {m + 2} } + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of Root of a x + b over x

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

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
Putting $n := \dfrac m 2$ and $m := -1$:
{{begin-eqn}}
{{eqn | l = \int \frac {\paren {\sqrt {a x + b} }^m} x \rd x
      | r = \int x^{-1} \paren {a x + b}^{m / 2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^0 \paren {a x + b}^{m / 2} } {-1 + \frac m 2 + 1} + \frac {\paren {\frac m 2} b} {-1 + \frac m 2 + 1} \int x^{-1} \paren {a x + b}^{m / 2 - 1} \rd x
      | c = 
}}
{{eqn | r = \frac {2 \paren {a x + b}^{m / 2} } m + b \int x^{-1} \paren {a x + b}^{\paren {m - 2} / 2} \rd x
      | c = simplifying
}}
{{eqn | r = \frac {2 \paren {\sqrt {a x + b} }^m } m + b \int \frac {\paren {\sqrt {a x + b} }^{m - 2} } x \rd x
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of Root of a x + b over x squared

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

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

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

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $a x + b$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$
Putting $n := -\dfrac m 2$ and $m := -1$:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {\sqrt {a x + b} }^m}
      | r = \int x^{-1} \paren {a x + b}^{-m/2}
      | c = 
}}
{{eqn | r = \frac {-x^0 \paren {a x + b}^{-m/2 + 1} } {\paren {-\frac m 2 + 1} b} + \frac {-1 - \frac m 2 + 2} {\paren {-\frac m 2 + 1} b} \int x^{-1} \paren {a x + b}^{- m/2 + 1} \rd x
      | c = 
}}
{{eqn | r = \frac 2 {\paren {m - 2} b \paren {a x + b}^{\paren {m - 2} / 2} } - \frac 1 b \int \frac {\d x} {x \paren {a x + b}^{\paren {m - 2} / 2} }
      | c = simplifying
}}
{{eqn | r = \frac 2 {\paren {m - 2} b \paren {\sqrt {a x + b} }^{m - 2} }  + \frac 1 b \int \frac {\d x} {x \paren {\sqrt {a x + b} }^{m - 2} }
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x + b by p x + q

:$\ds \int \frac {\d x} {\paren {a x + b} \paren {p x + q} } = \frac 1 {b p - a q} \ln \size {\frac {p x + q} {a x + b} } + C$
where $b p \ne a q$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a x + b} \paren {p x + q} }
      | r = \int \paren {\frac {-a} {\paren {b p - a q} \paren {a x + b} } + \frac p {\paren {b p - a q} \paren {p x + q} } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {b p - a q} \paren { {-a} \int \frac 1 {a x + b} \rd x + p \int \frac 1 {p x + q} \rd x}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac {-a} a \ln \size {a x + b} + \frac p p \ln \size {p x + q} } + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\ln \size {p x + q} - \ln \size {a x + b} } + C
      | c = simplification
}}
{{eqn | r = \frac 1 {b p - a q} \ln \size {\frac {p x + q} {a x + b} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over a x + b by p x + q

:$\ds \int \frac {x \rd x} {\paren {a x + b} \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac b a \ln \size {a x + b} - \frac q p \ln \size {p x + q} } + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {\paren {a x + b} \paren {p x + q} }
      | r = \int \paren {\frac b {\paren {b p - a q} \paren {a x + b} } - \frac q {\paren {b p - a q} \paren {p x + q} } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {b p - a q} \paren {b \int \frac 1 {a x + b} \rd x - q \int \frac 1 {p x + q} \rd x}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac b a \ln \size {a x + b} - \frac q p \ln \size {p x + q} } + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x + b squared by p x + q

:$\ds \int \frac {\d x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac 1 {a x + b} + \frac p {b p - a q} \ln \size {\frac {p x + q} {a x + b} } } + C$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\d x} {\paren {a x + b}^2 \paren {p x + q} }
      | c = 
}}
{{eqn | r = \int \paren {\frac 1 {b p - a q} \paren {\frac {-a p} {\paren {b p - a q} \paren {a x + b} } + \frac {-a} {\paren {a x + b}^2} + \frac {p^2} {\paren {b p - a q} \paren {p x + q} } } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac {-a p} {b p - a q} \int \frac {\d x} {a x + b} - a \int \frac {\d x} {\paren {a x + b}^2} + \frac {p^2} {b p - a q} \int \frac {\d x} {p x + q} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac {-a p} {b p - a q} \frac 1 a \ln \size {a x + b} - a \int \frac {\d x} {\paren {a x + b}^2} + \frac {p^2} {b p - a q} \frac 1 p \ln \size {p x + q} } + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac {-p} {b p - a q} \ln \size {a x + b} - a \frac {-1} {a \paren {a x + b} } + \frac p {b p - a q} \ln \size {p x + q} } + C
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac 1 {a x + b} + \frac p {b p - a q} \ln \size {\frac {p x + q} {a x + b} } } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over a x + b squared by p x + q

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

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} }
      | c = 
}}
{{eqn | r = \int \paren {\frac 1 {b p - a q} \paren {\frac {a q} {\paren {b p - a q} \paren {a x + b} } + \frac b {\paren {a x + b}^2} - \frac {p q} {\paren {b p - a q} \paren {p x + q} } } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac {a q} {b p - a q} \int \frac {\d x} {a x + b} + b \int \frac {\d x} {\paren {a x + b}^2} - \frac {p q} {b p - a q} \int \frac {\d x} {p x + q} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac {a q} {b p - a q} \frac 1 a \ln \size {a x + b} + b \int \frac {\d x} {\paren {a x + b}^2} + \frac {p q} {b p - a q} \frac 1 p \ln \size {p x + q} } + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {a x + b} + b \frac {-1} {a \paren {a x + b} } - \frac q {b p - a q} \ln \size {p x + q} } + C
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$
}}
{{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over a x + b squared by p x + q

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

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {x^2 \rd x} {\paren {a x + b}^2 \paren {p x + q} }
      | c = 
}}
{{eqn | r = \int \paren {\frac {-b^2} {a \paren {b p - a q} \paren {a x + b}^2} + \frac {q^2} {\paren {b p - a q}^2 \paren {p x + q} } + \frac {b \paren {b p - 2 a q} } {a \paren {b p - a q}^2 \paren {a x + b} } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac {-b^2} {a \paren {b p - a q} } \int \frac {\d x} {\paren {a x + b}^2} + \frac {q^2} {\paren {b p - a q}^2} \int \frac {\d x} {p x + q} + \frac {b \paren {b p - 2 a q} } {a \paren {b p - a q}^2} \int \frac {\d x} {a x + b}
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a x + b}^2}
      | r = \frac {-1} {a \paren {a x + b} }
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$
}}
{{eqn | l = \int \frac {\d x} {p x + q}
      | r = \frac 1 p \ln \size {p x + q}
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | l = \int \frac {\d x} {a x + b}
      | r = \frac 1 a \ln \size {a x + b}
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {x^2 \rd x} {\paren {a x + b}^2 \paren {p x + q} }
      | c = 
}}
{{eqn | r = \frac {-b^2} {a \paren {b p - a q} } \frac {-1} {a \paren {a x + b} } + \frac {q^2} {\paren {b p - a q}^2} \paren {\frac 1 p \ln \size {p x + q} } + \frac {b \paren {b p - 2 a q} } {a \paren {b p - a q}^2} \paren {\frac 1 a \ln \size {a x + b} } + C
      | c = Substitution
}}
{{eqn | r = \frac {b^2} {\paren {b p - a q} a^2 \paren {a x + b} } + \frac 1 {\paren {b p - a q}^2} \paren {\frac {q^2} p \ln \size {p x + q} + \frac {b \paren {b p - 2 a q} } {a^2} \ln \size {a x + b} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Friedrichs' Inequality

Let $G \subset \R^n$ be bounded domain.
Then for any $u \in \map {W^{1, 2}_0} G$:
:$\norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \norm {\nabla u}_{\map {L^2} G}$
where:
:$\map {\operatorname {diam} } G := \sup \limits_{x, y \mathop \in G} \size {x - y}$

Proof:
=== Smooth functions with compact support ===
{{explain|In several places in the below, this construct was seen: $\vert \nabla u (x_1, \dots, x_{m - 1}, t\vert^2$ where it appears that the close of the round bracket parenthesis "$)$" was been omitted. It has been assmued in the corrected version that it should be this: $\vert\nabla u (x_1,\dots,x_{m-1},t)\vert^2$ but it is not obvious and may require someone knowledgeable to check this out.}}
Let $u \in \map {C_0^\infty} G$.
Put $\map u x = 0$ if $x = \tuple {x_1, x_2, \ldots, x_n} \notin G$.
Then:
:$u \in \map {C_0^\infty} {\R^n}$
Denote by $a,b$ the extremes of the last coordinate: $a = \inf \limits_{x \mathop \in G} x_n$ and $b = \sup \limits_{y \mathop \in G} x_n$
Then for any $x$:
:$\ds \size {\map u {x_1, x_2, \ldots, x_n} }^2 = \size {\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \tuple {x_1, x_2, \ldots, x_{n - 1}, t} \rd t}^2$
By Cauchy-Bunyakovsky-Schwarz Inequality:
{{begin-eqn}}
{{eqn | l = \size {\int \limits_a^{x_n} \map {\frac {\partial u} {\partial x_n} } {x_1, \ldots, x_{n - 1}, t} \rd t}^2
      | o = \le
      | r = \int \limits_a^{x_n} 1^2 \rd t \int \limits_a^{x_n} \size {\map {\frac {\partial u} {\partial x_n} } {x_1, \ldots, x_{n - 1}, t} }^2 \rd t
      | c = 
}}
{{eqn | o = \le
      | r = \map {\operatorname {diam} } G \int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t
      | c = 
}}
{{end-eqn}}
Integrating this we get:
:$\ds \norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x$
By Fubini's Theorem:
{{begin-eqn}}
{{eqn | l = \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x
      | r = \int \limits_a^b \rd x_m \int \limits_{\mathbb R^n} \size {\map {\nabla u} {x_1, \ldots, x_{n - 1}, x_m} }^2 \rd x
      | c = 
}}
{{eqn | o = \le
      | r = \map {\operatorname {diam}^2} G \norm {\nabla u}^2_{\map {L^2} G}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of a x + b by Power of p x + q

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

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

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

Proof:
Aiming for an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
in order to use the technique of Integration by Parts, let:
{{begin-eqn}}
{{eqn | l = v
      | r = \paren {a x + b}^s
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = a s \paren {a x + b}^{s - 1}
      | c = Derivative of Power and Derivative of Function of Constant Multiple: Corollary
}}
{{end-eqn}}
In order to make $u \dfrac {\d v} {\d x}$ equal to the integrand, let:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac {\paren {a x + b}^{m - s + 1} } {a s} \paren {p x + q}^n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {a \paren {m - s + 1} \paren {a x + b}^{m - s} \paren {p x + q}^n + p n \paren {a x + b}^{m - s + 1} \paren {p x + q}^{n - 1} } {a s} 
      | c = Product Rule for Derivatives and above
}}
{{eqn | r = \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a s} \paren {a \paren {m - s + 1} \paren {p x + q} + p n \paren {a x + b} }
      | c = extracting common factor
}}
{{eqn | r = \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a s} \paren {a p x \paren {m - s + 1 + n} + a q \paren {m - s + 1} + p b n}
      | c = separating out terms in $x$
}}
{{end-eqn}}
Select $s$ such that $m - s + n + 1 = 0$, and so $s = m + n + 1$:
{{begin-eqn}}
{{eqn | l = \frac {\d u} {\d x}
      | r = \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a \paren {m + n + 1} } \paren {a q \paren {m - \paren {m + n + 1} + 1} + p b n}
      | c = term in $x$ vanishes
}}
{{eqn | r = \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a \paren {m + n + 1} } \paren {n \paren {b p - a q} }
      | c = simplifying
}}
{{end-eqn}}
Other instances of $s$ are left as they are, anticipating that they will cancel out later.
Thus:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \paren {a x + b}^m \paren {p x + q}^n \rd x
      | c = 
}}
{{eqn | r = \int \frac {\paren {a x + b}^{m - s + 1} } {a s} \paren {p x + q}^n a s \paren {a x + b}^{s - 1} \rd x
      | c = in the form $\ds \int u \frac {\d v} {\d x} \rd x$
}}
{{eqn | r = \frac {\paren {a x + b}^{m - s + 1} } {a s} \paren {p x + q}^n \paren {a x + b}^s
      | c = in the form $\ds u v - \int v \frac {\d u} {\d x} \rd x$
}}
{{eqn | o = 
      | ro= -
      | r = \int \paren {a x + b}^s \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a \paren {m + n + 1} } \paren {n \paren {p b - a q} } \rd x
      | c = 
}}
{{eqn | r = \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
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of a x + b by Power of p x + q/Increment of Power

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

Proof:
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$
Putting $n + 1$ for $n$, this yields:
{{begin-eqn}}
{{eqn | l = \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x
      | r = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} } {\paren {m + n + 2} a} - \frac {\paren {n + 1} \paren {b p - a q} } {\paren {m + n + 2} a} \int \paren {a x + b}^m \paren {p x + q}^n \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \paren {a x + b}^m \paren {p x + q}^n
      | r = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} } {\paren {n + 1} \paren {b p - a q} } - \frac {\paren {m + n + 2} a} {\paren {n + 1} \paren {b p - a q} } \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 {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - a \paren {m + n + 2} \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}
      | c = 
}}
{{end-eqn}}
{{qed}}