avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b

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

Proof:
Let $s \in \Z$. 
{{begin-eqn}}
{{eqn | l = v
      | r = x^s
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = s x^{s - 1}
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
Let $u \dfrac {\d v} {\d x} = x^m \paren {a x + b}^n$.
Then:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac {x^m \paren {a x + b}^n} {s x^{s - 1} }
      | c = from $(1)$
}}
{{eqn | r = \frac {x^{m - s + 1} \paren {a x + b}^n} s
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {x^{m - s + 1} a n \paren {a x + b}^{n - 1} } s + \frac {\paren {m - s + 1} x^{m - s} \paren {a x + b}^n} s
      | c = Primitive of Power of $a x + b$ and Product Rule for Derivatives
}}
{{eqn | r = \frac {x^{m - s} } s \paren {a x + b}^{n - 1} \paren {a n x + \paren {m - s + 1} \paren {a x + b} }
      | c = extracting $\dfrac {x^{m - s} \paren {a x + b}^n} s$ as a factor
}}
{{eqn | r = \frac {x^{m - s} } s \paren {a x + b}^{n - 1} \paren {a n x + a m x - a s x + a x + m b - s b + b}
      | c = multiplying out
}}
{{eqn | n = 2
      | r = \frac {x^{m - s} } s \paren {a x + b}^{n - 1} \paren {\paren {m + n + 1 - s} a x + \paren {m - s + 1} b}
      | c = gathering terms
}}
{{end-eqn}}
Let $s$ be selected such that $m + n + 1 - s = 0$.
Then $s = m + n + 1$.
Thus $(2)$ after rearrangement becomes:
:$\dfrac {\d u} {\d x} = \dfrac {x^{m - s} } {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - s + 1} b}$
Then:
{{begin-eqn}}
{{eqn | l = u v
      | r = \frac {x^{m - s + 1} \paren {a x + b}^n} {m + n + 1} x^s
      | c = 
}}
{{eqn | r = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1}
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = v \frac {\d u} {\d x}
      | r = x^s \dfrac {x^{m - s} } {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - s + 1} b}
      | c = 
}}
{{eqn | r = \dfrac {x^m} {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - \paren {m + n + 1} + 1} b}
      | c = 
}}
{{eqn | r = \dfrac {x^m} {m + n + 1} \paren {a x + b}^{n - 1} \paren {-n b}
      | c = 
}}
{{end-eqn}}
Thus by Integration by Parts:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
{{qed}}
Theorem: Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b

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

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 $a := 1, b := 0, p x + q := a x + b$:
{{begin-eqn}}
{{eqn | l = \int x^m \paren {a x + b}^n \rd x
      | r = \frac {\paren {1 x + 0}^{m + 1} \paren {a x + b}^n} {\paren {m + n + 1} 1} - \frac {n \paren {0 a - 1 b} } {\paren {m + n + 1} 1} \int \paren {1 x + 0}^m \paren {a x + b}^{n - 1} \rd x
      | c = 
}}
{{eqn | r = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} - \frac {n b} {\paren {m + n + 1} } \int x^m \paren {a x + b}^{n - 1} \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Distance from Subset to Element

Let $\struct {M, d}$ be a metric space.
Let $S \subseteq M$ be a subset of $M$.
Let $s \in S$.
Then:
:$\map d {s, S} = 0$
where $\map d {s, S}$ denotes the distance between $s$ and $S$.

Proof:
By Distance between Element and Subset is Nonnegative:
:$\map d {s, S} \ge 0$
Also, because $s \in S$, it follows that:
:$\map d {s, S} \le \map d {s, s} = 0$
Hence the result.
{{qed}}
Category:Distance Function
g2dc24h1w1u2grawyqcpwoyoz5ziw3y
Theorem: Primitive of Power of x by Power of a x + b/Decrement of Power of x

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

Proof:
Let $s \in \Z$. 
{{begin-eqn}}
{{eqn | l = v
      | r = \paren {a x + b}^s
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = a s \paren {a x + b}^{s - 1}
      | c = Power Rule for Derivatives and Derivatives of Function of $a x + b$
}}
{{end-eqn}}
Let $u \dfrac {\d v} {\d x} = x^m \paren {a x + b}^n$.
Then:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac {x^m \paren {a x + b}^n} {a s \paren {a x + b}^{s - 1} }
      | c = from $(1)$
}}
{{eqn | r = \frac {x^m} {a s} \paren {a x + b}^{n - s + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {m x^{m - 1} } {a s} \paren {a x + b}^{n - s + 1} + \frac {x^m} {a s} a \paren {n - s + 1} \paren {a x + b}^{n - s}
      | c = above rules and Product Rule for Derivatives
}}
{{eqn | r = \frac {x^{m - 1} } {a s} \paren {a x + b}^{n - s} \paren {m \paren {a x + b} + a \paren {n - s + 1} x}
      | c = extracting $\dfrac {x^{m - 1} } {a s} \paren {a x + b}^{n - s}$ as a factor
}}
{{eqn | n = 2
      | r = \frac {x^{m - 1} } {a s} \paren {a x + b}^{n - s} \paren {m b + a \paren {m + n - s + 1} x}
      | c = rearranging
}}
{{end-eqn}}
Let $s$ be selected such that $m + n + 1 - s = 0$.
Then $s = m + n + 1$.
Thus $(2)$ after rearrangement becomes:
:$\dfrac {\d u} {\d x} = \dfrac {m b x^{m - 1} \paren {a x + b}^{n - s} } {\paren {m + n + 1} a}$
Then:
{{begin-eqn}}
{{eqn | l = u v
      | r = \frac {x^m} {a s} \paren {a x + b}^{n - s + 1} \paren {a x + b}^s
      | c = 
}}
{{eqn | r = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a}
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = v \frac {\d u} {\d x}
      | r = \paren {a x + b}^s \frac {m b x^{m - 1} \paren {a x + b}^{n - s} } {\paren {m + n + 1} a}
      | c = 
}}
{{eqn | r = \frac {m b x^{m - 1} \paren {a x + b}^n} {\paren {m + n + 1} a}
      | c = 
}}
{{end-eqn}}
Thus by Integration by Parts:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
{{qed}}
Theorem: Primitive of Power of x by Power of a x + b/Decrement of Power of x

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

Proof:
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$
Setting $p := 1, q := 0, n := m, m := n$:
{{begin-eqn}}
{{eqn | l = \int x^m \paren {a x + b}^n \rd x
      | r = \frac {\paren {a x + b}^{n + 1} \paren {1 x + 0}^m} {\paren {m + n + 1} a} - \frac {m \paren {b 1 - a 0} } {\paren {m + n + 1} a} \int \paren {a x + b}^n \paren {1 x + 0}^{m - 1} \rd x
      | c = 
}}
{{eqn | r = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Distance from Subset to Supremum

Let $S \subseteq \R$ be a subset of the real numbers.
Suppose that the supremum $\sup S$ of $S$ exists.
Then:
:$\map d {\sup S, S} = 0$
where $\map d {\sup S, S}$ is the distance between $\sup S$ and $S$.

Proof:
By Distance between Element and Subset is Nonnegative:
:$\map d {\sup S, S} \ge 0$
By definition of supremum:
:$\forall \epsilon > 0: \exists s \in S: \map d {\sup S, s} < \epsilon$
meaning that, by nature of the infimum and the definition of $\map d {\sup S, S}$:
:$\forall \epsilon > 0: \map d {\sup S, S} < \epsilon$
Together, these two observations lead to the conclusion that:
:$\map d {\sup S, S} = 0$
as desired.
{{qed}}
Theorem: Distance between Element and Subset is Nonnegative

Let $\struct {M, d}$ be a metric space.
Let $x \in M$ and $S \subseteq M$.
Then:
:$\map d {x, S} \ge 0$
where $\map d {x, S}$ is the distance between $x$ and $S$.

Proof:
By definition of the distance between $x$ and $S$:
:$\map d {x, S} = \ds \inf_{s \mathop \in S} \map d {x, s}$
From the metric space axioms:
:$\forall s \in M: \map d {x, s} \ge 0$
Hence by the nature of the infimum:
:$\map d {x, S} \ge 0$
as desired.
{{qed}}
Category:Distance Function
70b2wntin5ezf0alcdgs8v9lmbkscm4
Theorem: Distance from Subset to Infimum

Let $S \subseteq \R$ be a subset of the real numbers.
Suppose that the infimum $\inf S$ of $S$ exists.
Then:
:$\map d {\inf S, S} = 0$
where $\map d {\inf S, S}$ is the distance between $\inf S$ and $S$.

Proof:
By Distance between Element and Subset is Nonnegative:
:$\map d {\inf S, S} \ge 0$
By definition of infimum:
:$\forall \epsilon > 0: \exists s \in S: \map d {\inf S, s} < \epsilon$
meaning that, by nature of the infimum and the definition of $\map d {\inf S, S}$:
:$\forall \epsilon > 0: \map d {\inf S, S} < \epsilon$
Together, these two observations lead to the conclusion that:
:$\map d {\inf S, S} = 0$
as desired.
{{qed}}
Category:Real Analysis
ge783khyr89w1rtnt8t2t2aw56bn4vm
Theorem: Primitive of Power of x by Power of a x + b/Increment of Power of a x + b

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

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
Substituting $n + 1$ for $n$:
{{begin-eqn}}
{{eqn | l = \int x^m \paren {a x + b}^{n + 1} \rd x
      | r = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \frac {\paren {n + 1} b} {m + n + 2} \int x^m \paren {a x + b}^n \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\paren {n + 1} b} {m + n + 2} \int x^m \paren {a x + b}^n \rd x
      | r = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \int x^m \paren {a x + b}^{n + 1} \rd x
      | c = rearrangement
}}
{{eqn | ll= \leadsto
      | l = \int x^m \paren {a x + b}^n \rd x
      | r = \frac {m + n + 2} {\paren {n + 1} b} \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x
      | c = rearrangement
}}
{{eqn | ll= \leadsto
      | l = \int x^m \paren {a x + b}^n \rd x
      | r = \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
      | c = rearrangement
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Power of a x + b/Increment of Power of a x + b

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

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 $a := 1, b := 0, p x + q := a x + b$:
{{begin-eqn}}
{{eqn | l = \int x^m \paren {a x + b}^n \rd x
      | r = \frac 1 {\paren {n + 1} \paren {0 a - 1 b} } \paren {\paren {1 x + 0}^{m + 1} \paren {a x + b}^{n + 1} - 1 \paren {m + n + 2} \int \paren {1 x + 0}^m \paren {a x + b} ^{n + 1} \rd x}
      | c = 
}}
{{eqn | r = \frac 1 {\paren {n + 1} \paren {-b} } \paren {x^{m + 1} \paren {a x + b}^{n + 1} - \paren {m + n + 2} \int x^m \paren {a x + b}^{n + 1} \rd x}
      | c = 
}}
{{eqn | r = \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
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Power of a x + b/Increment of Power of x

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

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Substituting $m + 1$ for $m$:
{{begin-eqn}}
{{eqn | l = \int x^{m + 1} \paren {a x + b}^n \rd x
      | r = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + n + 2} a} - \frac {\paren {m + 1} b} {\paren {m + n + 2} a} \int x^m \paren {a x + b}^n \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\paren {m + 1} b} {\paren {m + n + 2} a} \int x^m \paren {a x + b}^n \rd x
      | r = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + n + 2} a} - \int x^{m + 1} \paren {a x + b}^n \rd x
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \int x^m \paren {a x + b}^n \rd x
      | r = \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + n + 2} a} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x
      | c = rearranging
}}
{{eqn | r = \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
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Power of a x + b/Increment of Power of x

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

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 $p := 1, q := 0, m := n, n := m$:
{{begin-eqn}}
{{eqn | l = \int x^m \paren {a x + b}^n \rd x
      | r = \frac 1 {\paren {m + 1} \paren {b 1 - a 0} } \paren {\paren {a x + b}^{n + 1} \paren {1 x + 0}^{m + 1} - a \paren {m + n + 2} \int \paren {a x + b}^n \paren {0 x + 1}^{m + 1} \rd x}
      | c = 
}}
{{eqn | r = \frac 1 {\paren {m + 1} b} \paren {\paren {a x + b}^{n + 1} x^{m + 1} - a \paren {m + n + 2} \int \paren {a x + b}^n x^{m + 1} \rd x}
      | c = 
}}
{{eqn | r = \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
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Root of a x + b

:$\ds \int \frac {\d x} {\sqrt{a x + b} } = \frac {2 \sqrt {a x + b} } a + C$
where $a x + b > 0$.

Proof:
Put $u = \sqrt{a x + b}$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\sqrt {a x + b} }
      | r = \frac 2 a \int u \frac {\d u} u
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 a \int \rd u
      | c = 
}}
{{eqn | r = \frac 2 a u + C
      | c = Primitive of Constant
}}
{{eqn | r = \frac {2 \sqrt {a x + b} } a + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Root of a x + b

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \sqrt {a x + b}
      | c = 
}}
{{eqn | l = x
      | r = \frac {u^2 - b} a
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \map F {\sqrt {a x + b} }
      | r = \frac x {\sqrt {a x + b} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map F u
      | r = \paren {\frac {u^2 - b} a} \frac 1 u
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {\sqrt {a x + b} }
      | r = \frac 2 a \int u \paren {\frac {u^2 - b} a} \frac 1 u \rd u
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 {a^2} \int \paren {u^2 - b} \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 2 {a^2} \paren {\frac {u^3} 3 - b u} + C
      | c = Primitive of Power and Primitive of Constant
}}
{{eqn | r = \frac 2 {a^2} \paren {\frac {\paren {a x + b} \sqrt {a x + b} } 3 - b \sqrt {a x + b} } + C
      | c = substituting for $u$
}}
{{eqn | r = \frac 2 {3 a^2} \paren {\paren {a x + b} - 3 b} \sqrt {a x + b} + C
      | c = extracting common factors
}}
{{eqn | r = \frac {2 \paren {a x - 2 b} \sqrt {a x + b} } {3 a^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over Root of a x + b

:$\ds \int \frac {x^2 \rd x} {\sqrt {a x + b} } = \frac {2 \paren {3 a^2 x^2 - 4 a b x + 8 b^2} \sqrt {a x + b} } {15 a^3}$

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}}
Thus:
{{begin-eqn}}
{{eqn | l = \map F {\sqrt {a x + b} }
      | r = \frac {x^2} {\sqrt {a x + b} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map F u
      | r = \paren {\frac {u^2 - b} a}^2 \frac 1 u
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\sqrt {a x + b} }
      | r = \frac 2 a \int u \paren {\frac {u^2 - b} a}^2 \frac 1 u \rd u
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 {a^3} \int \paren {u^4 - 2 b u^2 + b^2} \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 2 {a^3} \paren {\frac {u^5} 5 - \frac {2 b u^3} 3 + b^2 u} + C
      | c = Primitive of Power and Primitive of Constant
}}
{{eqn | r = \frac 2 {a^3} \paren {\frac {\paren {a x + b}^2} 5 - \frac {2 b \paren {a x + b} } 3 + b^2} \sqrt {a x + b} + C
      | c = substituting for $u$ and extracting common factors
}}
{{eqn | r = \frac 2 {15 a^2} \paren {3 a^2 x^2 + 6 a b x + 3 b^2 - 10 a b x - 10 b^2 + 15 b^2} \sqrt {a x + b} + C
      | c = multiplying out and combining fractions
}}
{{eqn | r = \frac {2 \paren {3 a^2 x^2 - 4 a b x + 8 b^2} \sqrt {a x + b} } {15 a^3} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Cotangent Form

Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $\size x > a$.
Then:
:$\ds \int \frac {\d x} {x^2 - a^2} = -\frac 1 a \coth^{-1} {\frac x a} + C$

Proof:
Let $\size x > a$.
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \coth^{-1} {\frac x a}
      | c = {{Defof|Real Inverse Hyperbolic Cotangent}}, which is defined where $\size {\dfrac x a} > 1$
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \coth u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = -a \csch^2 u
      | c = Derivative of Hyperbolic Cotangent
}}
{{eqn | ll= \leadsto
      | l = \int \frac 1 {x^2 - a^2} \rd x
      | r = \int \frac {-a \csch^2 u} {a^2 \coth^2 u - a^2} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = -\frac a {a^2} \int \frac {\csch^2 u} {\coth^2 u - 1} \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 a \int \frac {\csch^2 u} {\csch^2 u} \rd u
      | c = Difference of Squares of Hyperbolic Cotangent and Cosecant
}}
{{eqn | r = -\frac 1 a \int \rd u
}}
{{eqn | r = -\frac 1 a u + C
      | c = Integral of Constant
}}
{{eqn | r = -\frac 1 a \coth^{-1} {\frac x a} + C
      | c = Definition of $u$
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal of x by Root of a x + b

For $a > 0$ and for $x \ne 0$:
:<nowiki>$\ds \int \frac {\d x} {x \sqrt {a x + b} } = \begin {cases}
\dfrac 1 {\sqrt b} \ln \size {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C & : b > 0 \\ \\
\dfrac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C & : b < 0 \end {cases}$</nowiki>
where $a x + b > 0$.

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}}
Thus:
{{begin-eqn}}
{{eqn | l = \map F {\sqrt {a x + b} }
      | r = \frac 1 {x \sqrt {a x + b} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map F u
      | r = \paren {\frac a {u^2 - b} } \frac 1 u
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \sqrt {a x + b} }
      | r = \frac 2 a \int u \paren {\frac a {u^2 - b} } \frac 1 u \rd u
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = 2 \int \frac {\d u} {u^2 - b}
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Let $b > 0$.
Let $d = \sqrt b$.
Then:
{{begin-eqn}}
{{eqn | l = 2 \int \frac {\d u} {u^2 - b}
      | r = 2 \int \frac {\d u} {u^2 - d^2}
      | c = 
}}
{{eqn | r = 2 \paren {\frac 1 {2 d} } \ln \size {\frac {u - d} {u + d} } + C
      | c = Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form
}}
{{eqn | r = \dfrac 1 {\sqrt b} \ln \size {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C
      | c = substituting for $u$ and $d$
}}
{{end-eqn}}
{{qed|lemma}}
Let $b < 0$.
Let $d = \sqrt {-b}$.
Then:
{{begin-eqn}}
{{eqn | l = 2 \int \frac {\d u} {u^2 - b}
      | r = 2 \int \frac {\d u} {u^2 - \paren {-d^2} }
      | c = 
}}
{{eqn | r = 2 \int \frac {\d u} {u^2 + d^2}
      | c = 
}}
{{eqn | r = 2 \paren {\frac 1 d} \arctan {\frac u d} + C
      | c = Primitive of Reciprocal of $x^2 + a^2$: Arctangent Form
}}
{{eqn | r = \frac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C
      | c = substituting for $u$ and $d$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by Root of a x + b

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

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

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

Proof:
Let $u = \sqrt{a x + b}$.
Then:
{{begin-eqn}}
{{eqn | l = \int \sqrt {a x + b} \rd x
      | r = \frac 2 a \int u^2 \rd u
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 a \frac {u^3} 3
      | c = Primitive of Power
}}
{{eqn | r = \frac {2 \sqrt {\paren {a x + b}^3} } {3 a}
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a x + b

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

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 $m = -1$ and $n = \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } x \rd x
      | r = \int x^{-1} \paren {a x + b}^{1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^0 \paren {a x + b}^{1/2} } {\frac 1 2} + \frac {\frac 1 2 b} {\frac 1 2} \int x^{-1} \paren {a x + b}^{- 1/2} \rd x
      | c = 
}}
{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}