avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of x by Inverse Hyperbolic Cotangent of x over a

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

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcoth \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac a {a^2 - x^2}
      | c = Derivative of $\arcoth \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \arcoth \frac x a \rd x
      | r = \frac {x^2} 2 \arcoth \frac x a - \int \frac {x^2} 2 \paren {\frac a {a^2 - x^2} } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arcoth \frac x a + \frac a 2 \int \frac {x^2 \rd x} {x^2 - a^2} + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^2} 2 \arcoth \frac x a + \frac a 2 \paren {x - a \arcoth \frac x a} + C
      | c = Primitive of $\dfrac {x^2} {x^2 - a^2}$
}}
{{eqn | r = \frac {a x} 2 + \frac {x^2 - a^2} 2 \arcoth \frac x a + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Inverse Hyperbolic Secant of x over a

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

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arsech \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {-a} {x \sqrt {a^2 - x^2} }
      | c = Derivative of $\arsech \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \arsech \frac x a \rd x
      | r = \frac {x^2} 2 \arsech \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {a^2 - x^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arsech \frac x a + \frac a 2 \int \frac {x \rd x} {\sqrt {a^2 - x^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {x^2} 2 \arsech \frac x a + \frac a 2 \paren {-\sqrt {a^2 - x^2} } + C
      | c = Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$
}}
{{eqn | r = \frac {x^2} 2 \arsech \frac x a - \frac {a \sqrt {a^2 - x^2} } 2 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Inverse Hyperbolic Cosecant of x over a

:<nowiki>$\ds \int x \arcsch \frac x a \rd x = \begin {cases}
\dfrac {x^2} 2 \arcsch \dfrac x a + \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x > 0 \\ \\
\dfrac {x^2} 2 \arcsch \dfrac x a - \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x < 0
\end {cases}$</nowiki>

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsch \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \dfrac {-a} {\size x \sqrt {a^2 + x^2} }
      | c = Derivative of $\arcsch \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \arcsch \frac x a \rd x
      | r = \frac {x^2} 2 \arcsch \frac x a - \int \frac {x^2} 2 \paren {\dfrac {-a} {\size x \sqrt {a^2 + x^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arcsch \frac x a + \frac a 2 \int \frac {x^2 \rd x} {\size x \sqrt {a^2 + x^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {x^2} 2 \arcsch \frac x a \begin {cases} \mathop + \dfrac a 2 \int \dfrac {x \rd x} {\sqrt {x^2 + a^2} } + C & : x > 0 \\ \mathop - \dfrac a 2 \int \dfrac {x \rd x} {\sqrt {x^2 + a^2} } + C & : x < 0 \end {cases}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = \frac {x^2} 2 \arcsch \frac x a \begin {cases} \mathop + \dfrac a 2 \sqrt {x^2 + a^2} + C & : x > 0 \\ \mathop - \dfrac a 2 \sqrt {x^2 + a^2} + C & : x < 0 \end {cases}
      | c = Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$
}}
{{eqn | r = <nowiki>\begin {cases}
\dfrac {x^2} 2 \arcsch \dfrac x a + \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x > 0 \\ \\
\dfrac {x^2} 2 \arcsch \dfrac x a - \dfrac {a \sqrt {x^2 + a^2} } 2 + C & : x < 0 
\end {cases}</nowiki>
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Inverse Hyperbolic Sine of x over a

:$\ds \int x^2 \arsinh \frac x a \rd x = \frac {x^3} 3 \arsinh \frac x a + \frac {\paren {2 a^2 - x^2} \sqrt {x^2 + a^2} } 9 + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arsinh \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 {\sqrt {x^2 + a^2} }
      | c = Derivative of $\arsinh \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^3} 3
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \arsinh \frac x a \rd x
      | r = \frac {x^3} 3 \arsinh \frac x a - \int \frac {x^3} 3 \paren {\frac 1 {\sqrt {x^2 + a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \arsinh \frac x a - \frac 1 3 \int \frac {x^3 \rd x} {\sqrt {x^2 + a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {x^3} 3 \arsinh \frac x a - \frac 1 3 \paren {\frac {\paren {\sqrt {x^2 + a^2} }^3} 3 - a^2 \sqrt {x^2 + a^2} } + C
      | c = Primitive of $\dfrac {x^3} {\sqrt {x^2 + a^2} }$
}}
{{eqn | r = \frac {x^3} 3 \arsinh \frac x a + \frac {\paren {2 a^2 - x^2} \sqrt {x^2 + a^2} } 9 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Inverse Hyperbolic Cosine of x over a

:$\ds \int x^2 \arcosh \frac x a \rd x = \dfrac {x^3} 3 \arcosh \dfrac x a - \dfrac {\paren {x^2 + 2 a^2} \sqrt {x^2 - a^2} } 9 + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcosh \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 {\sqrt {x^2 - a^2} }
      | c = Derivative of $\arcosh \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^3} 3
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \arcosh \frac x a \rd x
      | r = \frac {x^3} 3 \arcosh \frac x a - \int \frac {x^3} 3 \paren {\frac 1 {\sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \arcosh \frac x a - \frac 1 3 \int \frac {x^3 \rd x} {\sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {x^3} 3 \arcosh \frac x a - \frac 1 3 \paren {\frac {\paren {\sqrt {x^2 - a^2} }^3} 3 + a^2 \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac {x^3} {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {x^3} 3 \arcosh \frac x a - \frac {\paren {x^2 + 2 a^2} \sqrt {x^2 - a^2} } 9 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Inverse Hyperbolic Tangent of x over a

:$\ds \int x^2 \artanh \frac x a \rd x = \frac {a x^2} 6 + \frac {x^3} 3 \artanh \frac x a + \frac {a^3} 6 \map \ln {a^2 - x^2} + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \artanh \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac a {a^2 - x^2}
      | c = Derivative of $\artanh \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^3} 3
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \artanh \frac x a \rd x
      | r = \frac {x^3} 3 \artanh \frac x a - \int \frac {x^3} 3 \paren {\frac a {a^2 - x^2} } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \artanh \frac x a - \frac a 3 \int \frac {x^3 \rd x} {a^2 - x^2} + C
      | c = simplifying
}}
{{eqn | r = \frac {x^3} 3 \artanh \frac x a - \frac a 3 \paren {-\frac {x^2} 2 - \frac {a^2} 2 \map \ln {a^2 - x^2} } + C
      | c = Primitive of $\dfrac {x^3} {a^2 - x^2}$
}}
{{eqn | r = \frac {a x^2} 6 + \frac {x^3} 3 \artanh \frac x a + \frac {a^3} 6 \map \ln {a^2 - x^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Inverse Hyperbolic Cotangent of x over a

:$\ds \int x^2 \arcoth \frac x a \rd x = \frac {a x^2} 6 + \frac {x^3} 3 \arcoth \frac x a + \frac {a^3} 6 \map \ln {x^2 - a^2} + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcoth \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {-a} {x^2 - a^2}
      | c = Derivative of $\arcoth \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^3} 3
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^2 \arcoth \frac x a \rd x
      | r = \frac {x^3} 3 \arcoth \frac x a - \int \frac {x^3} 3 \paren {\frac {-a} {x^2 - a^2} } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^3} 3 \arcoth \frac x a + \frac a 3 \int \frac {x^3 \rd x} {x^2 - a^2} + C
      | c = simplifying
}}
{{eqn | r = \frac {x^3} 3 \arcoth \frac x a + \frac a 3 \paren {\frac {x^2} 2 + \frac {a^2} 2 \map \ln {x^2 - a^2} } + C
      | c = Primitive of $\dfrac {x^3} {x^2 - a^2}$
}}
{{eqn | r = \frac {a x^2} 6 + \frac {x^3} 3 \arcoth \frac x a + \frac {a^3} 6 \map \ln {x^2 - a^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Inverse Hyperbolic Cosine of x over a over x

{{begin-eqn}}
{{eqn | l = \int \dfrac 1 x \arcosh \dfrac x a \rd x
      | r = \dfrac 1 2 \ln^2 \paren {\dfrac {2 x} a} + \sum_{n \mathop \ge 1} \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac a x}^{2 n} + C
      | c = 
}}
{{eqn | r = \dfrac 1 2 \ln^2 \paren {\dfrac {2 x} a} + \dfrac 1 {2 \times 2^2} \paren {\dfrac a x}^2 + \dfrac {1 \times 3} {2 \times 4 \times 4^2} \paren {\dfrac a x}^4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6^2} \paren {\dfrac a x}^6 + \dotsb + C
      | c = 
}}
{{end-eqn}}

Proof:
For $\arcosh \dfrac x a > 0$:
{{begin-eqn}}
{{eqn | l = \arcosh \dfrac x a
      | r = \ln \frac {2 x} a - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \paren {\frac a x}^{2 n} }
      | c = Power Series Expansion for Real Area Hyperbolic Cosine
}}
{{eqn | ll= \leadsto
      | l = \frac 1 x \arcosh \dfrac x a
      | r = \dfrac 1 x \ln \frac {2 x} a - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } a^{2 n} \paren {\frac 1 x}^{2 n + 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac 1 x \arcosh \dfrac x a \rd x
      | r = \int \paren {\dfrac 1 x \ln \frac {2 x} a - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } a^{2 n} \paren {\frac 1 x}^{2 n + 1} } } \rd x
      | c = 
}}
{{eqn | r = \int \frac 1 x \ln \frac {2 x} a \rd x - \paren {\sum_{n \mathop = 1}^\infty \int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } a^{2 n} \paren {\frac 1 x}^{2 n + 1} \rd x}
      | c = Fubini's Theorem
}}
{{eqn | r = \frac 1 2 \map {\ln^2} {\dfrac {2 x} a} - \paren {\sum_{n \mathop = 1}^\infty \int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } a^{2 n} \paren {\frac 1 x}^{2 n + 1} \rd x} + C
      | c = Primitive of $\dfrac {\ln x} x$: Corollary
}}
{{eqn | r = \frac 1 2 \map {\ln^2} {\dfrac {2 x} a} - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 n \paren {-2 n} } a^{2 n} \paren {\dfrac 1 x}^{2 n} } + C
      | c = Primitive of Power
}}
{{eqn | r = \frac 1 2 \map {\ln^2} {\dfrac {2 x} a} + \sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac a x}^{2 n} + C
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Inverse Hyperbolic Secant of x over a over x

{{begin-eqn}}
{{eqn | l = \int \dfrac 1 x \arsech \dfrac x a \rd x
      | r = -\frac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} - \sum_{n \mathop \ge 1} \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac x a}^{2 n} + C
      | c = 
}}
{{eqn | r = -\dfrac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} - \dfrac 1 {2 \times 2^2} \paren {\dfrac x a}^2 - \dfrac {1 \times 3} {2 \times 4 \times 4^2} \paren {\dfrac x a}^4 - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6^2} \paren {\dfrac x a}^6 + C
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \arsech \dfrac x a
      | r = \ln \frac {2 a} x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \paren {\frac x a}^{2 n} }
      | c = Power Series Expansion for Real Area Hyperbolic Secant
}}
{{eqn | ll= \leadsto
      | l = \frac 1 x \arsech \dfrac x a
      | r = \dfrac 1 x \ln \frac {2 a} x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac 1 x \arsech \dfrac x a \rd x
      | r = \int \paren {\dfrac 1 x \ln \frac {2 a} x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} } } \rd x
      | c = 
}}
{{eqn | r = \int -\frac 1 x \ln \frac x {2 a} \rd x - \paren {\sum_{n \mathop = 1}^\infty \int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} \rd x}
      | c = Fubini's Theorem
}}
{{eqn | r = -\frac 1 2 \map {\ln^2} {\dfrac x {2 a} } - \paren {\sum_{n \mathop = 1}^\infty \int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} \rd x} + C
      | c = Primitive of $\dfrac {\ln x} x$: Corollary
}}
{{eqn | r = -\frac 1 2 \map {\ln^2} {\dfrac x {2 a} } - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 n \paren {2 n} } \frac 1 {a^{2 n} } \dfrac {x^{2 n} } {2 n} } + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} + \map {\ln^2} 2 - \sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac x a}^{2 n} + C
      | c = rearranging
}}
{{eqn | r = -\frac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} - \sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac x a}^{2 n} + C
      | c = subsuming $\map {\ln^2} 2$ into the constant of integration
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Inverse Hyperbolic Sine of x over a over x squared

:$\ds \int \frac 1 {x^2} \arsinh \dfrac x a \rd x = -\frac 1 x \arsinh \dfrac x a - \frac 1 a \map \ln {\frac a x + \frac {\sqrt {a^2 + x^2} } {\size x} }$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arsinh \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 {\sqrt {x^2 + a^2} }
      | c = Derivative of $\arsinh \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} x
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac 1 {x^2} \arsinh \dfrac x a \rd x
      | r = \paren {\arsinh \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac 1 {\sqrt {x^2 + a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = -\frac 1 x \arsinh \dfrac x a + \int \frac {\d x} {x \sqrt {x^2 + a^2} } + C
      | c = simplifying
}}
{{eqn | r = -\frac 1 x \arsinh \dfrac x a - \frac 1 a \map \ln {\frac a x + \frac {\sqrt {a^2 + x^2} } {\size x} } + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Inverse Hyperbolic Cosine of x over a over x squared

:$\ds \int \frac 1 {x^2} \arcosh \dfrac x a \rd x = -\frac 1 x \arcosh \dfrac x a + \frac 1 a \arcsec \size {\frac x a} + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcosh \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 {\sqrt {x^2 - a^2} }
      | c = Derivative of $\arcosh \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} x
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac 1 {x^2} \arcosh \dfrac x a \rd x
      | r = \paren {\arcosh \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac 1 {\sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = -\frac 1 x \arcosh \dfrac x a + \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = -\frac 1 x \arcosh \dfrac x a + \frac 1 a \arcsec \size {\frac x a} + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Inverse Hyperbolic Tangent of x over a over x squared

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

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \artanh \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac a {a^2 - x^2}
      | c = Derivative of $\artanh \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} x
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac 1 {x^2} \artanh \dfrac x a \rd x
      | r = \paren {\artanh \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac a {a^2 - x^2} } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac 1 x \artanh \dfrac x a + a \int \frac {\d x} {x \paren {a^2 - x^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 x \artanh \dfrac x a + a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2} {a^2 - x^2} } } + C
      | c = Primitive of $\dfrac 1 {x \paren {a^2 - x^2} }$
}}
{{eqn | r = \frac 1 x \artanh \dfrac x a + \frac 1 {2 a} \map \ln {\frac {x^2} {a^2 - x^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Inverse Hyperbolic Cotangent of x over a over x squared

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

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcoth \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {-a} {x^2 - a^2}
      | c = Derivative of $\arcoth \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} x
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac 1 {x^2} \arcoth \dfrac x a \rd x
      | r = \paren {\arcoth \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac {-a} {x^2 - a^2} } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = -\frac 1 x \arcoth \dfrac x a - a \int \frac {\d x} {x \paren {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = -\frac 1 x \arcoth \dfrac x a - a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } } + C
      | c = Primitive of $\dfrac 1 {x \paren {x^2 - a^2} }$
}}
{{eqn | r = -\frac 1 x \arcoth \dfrac x a - \frac 1 {2 a} \map \ln {\frac {x^2 - a^2} {x^2} } + C
      | c = simplifying
}}
{{eqn | r = -\frac 1 x \arcoth \dfrac x a + \frac 1 {2 a} \map \ln {\frac {x^2} {x^2 - a^2} } + C
      | c = Logarithm of Reciprocal
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Inverse Hyperbolic Sine of x over a

:$\ds \int x^m \sinh^{-1} \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \sinh^{-1} \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} } {\sqrt {x^2 + a^2} } \rd x + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \sinh^{-1} \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 {\sqrt {x^2 + a^2} }
      | c = Derivative of $\sinh^{-1} \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^{m + 1} } {m + 1}
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \sinh^{-1} \frac x a \rd x
      | r = \paren {\sinh^{-1} \frac x a} \paren {\frac {x^{m + 1} } {m + 1} } - \int \paren {\frac {x^{m + 1} } {m + 1} } \paren {\frac 1 {\sqrt {x^2 + a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \sinh^{-1} \frac x a - \frac 1 {m + 1}  \int \frac {x^{m + 1} } {\sqrt {x^2 + a^2} } \rd x + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Inverse Hyperbolic Cosine of x over a

:$\ds \int x^m \arcosh \frac x a \rd x = \dfrac {x^{m + 1} } {m + 1} \arcosh \dfrac x a - \dfrac 1 {m + 1} \int \dfrac {x^{m + 1} } {\sqrt {x^2 - a^2} } \rd x + C$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcosh \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 {\sqrt {x^2 - a^2} }
      | c = Derivative of $\arcosh \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^{m + 1} } {m + 1}
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arcosh \frac x a \rd x
      | r = \paren {\arcosh \frac x a} \paren {\frac {x^{m + 1} } {m + 1} } - \int \paren {\frac {x^{m + 1} } {m + 1} } \paren {\frac 1 {\sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arcosh \frac x a - \frac 1 {m + 1}  \int \frac {x^{m + 1} } {\sqrt {x^2 - a^2} } \rd x + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}