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

:<nowiki>$\ds \int \arcsch \frac x a \rd x = \begin {cases}
x \arcsch \dfrac x a + a \arsinh \dfrac x a + C & : x > 0 \\ \\
x \arcsch \dfrac x a - a \arsinh \dfrac x a + C & : x < 0
\end {cases}$</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 = \frac {-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 = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = x
      | c = Primitive of Constant
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \arcsch \frac x a \rd x
      | r = x \arcsch \frac x a - \int x \paren {\frac {-a} {\size x \sqrt {a^2 + x^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = x \arcsch \frac x a + a \int \frac {x \rd x} {\size x \sqrt {a^2 + x^2} } + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x \arcsch \frac x a \begin {cases} \mathop + a \ds \int \dfrac {\d x} {\sqrt {a^2 + x^2} } + C & : x > 0 \\ \\ \mathop - a \ds \int \dfrac {\d x} {\sqrt {a^2 + x^2} } + C & : x < 0 \end {cases}
      | c = {{Defof|Absolute Value}} 
}}
{{eqn | r = \begin {cases} x \arcsch \dfrac x a + a \arsinh \dfrac x a + C & : x > 0 \\ \\ x \arcsch \dfrac x a - a \arsinh \dfrac x a + C & : x < 0 \end {cases}
      | c = Primitive of $\dfrac 1 {\sqrt {a^2 + x^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Inverse Hyperbolic Sine of x over a

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

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
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \arsinh \frac x a \rd x
      | r = \frac {x^2} 2 \arsinh \frac x a - \int \frac {x^2} 2 \paren {\frac 1 {\sqrt {x^2 + a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arsinh \frac x a - \frac 1 2 \int \frac {x^2 \rd x} {\sqrt {x^2 + a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {x^2} 2 \arsinh \frac x a - \frac 1 2 \paren {\frac {x \sqrt {x^2 + a^2} } 2 - \frac {a^2} 2 \arsinh \frac x a} + C
      | c = Primitive of $\dfrac {x^2} {\sqrt {x^2 + a^2} }$
}}
{{eqn | r = \paren {\frac {x^2} 2 + \frac {a^2} 4} \arsinh \frac x a - \frac {x \sqrt {x^2 + a^2} } 4 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Real Area Hyperbolic Cosine of x over a in Logarithm Form

:$\arcosh \dfrac x a = \map \ln {x + \sqrt {x^2 - a^2} } - \ln a$

Proof:
{{begin-eqn}}
{{eqn | l = \arcosh  \frac x a
      | r = \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - 1} }
      | c = {{Defof|Real Area Hyperbolic Cosine}}
}}
{{eqn | r = \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } }
      | c = 
}}
{{eqn | r = \map \ln {\frac x a + \frac {\sqrt {x^2 - a^2} } a}
      | c = 
}}
{{eqn | r = \map \ln {\frac {x + \sqrt {x^2 - a^2} } a}
      | c = 
}}
{{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } - \ln a
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Real Area Hyperbolic Tangent of x over a in Logarithm Form

:$\artanh \dfrac x a = \dfrac 1 2 \map \ln {\dfrac {a + x} {a - x} }$

Proof:
{{begin-eqn}}
{{eqn | l = \artanh  \frac x a
      | r = \frac 1 2 \map \ln {\frac {1 + \frac x a} {1 - \frac x a} }
      | c = {{Defof|Real Area Hyperbolic Tangent}}
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {a + x} {a - x} }
      | c = multiplying top and bottom by $a$
}}
{{end-eqn}}
{{qed}}
Theorem: Real Area Hyperbolic Cotangent of x over a in Logarithm Form

:$\arcoth \dfrac x a = \dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} }$

Proof:
{{begin-eqn}}
{{eqn | l = \arcoth \frac x a
      | r = \frac 1 2 \map \ln {\frac {\frac x a + 1} {\frac x a - 1} }
      | c = {{Defof|Real Area Hyperbolic Cotangent}}
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {x + a} {x - a} }
      | c = multiplying top and bottom by $a$
}}
{{end-eqn}}
{{qed}}
Theorem: Real Area Hyperbolic Secant of x over a in Logarithm Form

:$\arsech \dfrac x a = \map \ln {\dfrac {a + \sqrt {a^2 - x^2} } x}$

Proof:
{{begin-eqn}}
{{eqn | l = \arsech \frac x a
      | r = \map \ln {\frac {1 + \sqrt {1 - \paren {\frac x a}^2} } {\frac x a} }
      | c = {{Defof|Real Area Hyperbolic Secant}}
}}
{{eqn | r = \map \ln {\frac {a \paren {1 + \sqrt {\dfrac {a^2 - x^2} {a^2} } } } x}
      | c = 
}}
{{eqn | r = \map \ln {\frac {a \paren {\dfrac {a + \sqrt {a^2 - x^2} } a} } x}
      | c = 
}}
{{eqn | r = \map \ln {\frac {a + \sqrt {a^2 - x^2} } x}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Real Area Hyperbolic Cosecant of x over a in Logarithm Form

For $a > 0$:
:$\arcsch \dfrac x a = \map \ln {\dfrac {a + \sqrt {a^2 + x^2} } {\size x} }$

Proof:
We have that $\arcsch \dfrac x a$ is defined whenever $x \ne 0$.
{{begin-eqn}}
{{eqn | l = \arcsch \frac x a
      | r = \map \ln {\frac 1 {x/a} + \frac {\sqrt {1 + \paren {\frac x a}^2} } {\size {\frac x a} } }
      | c = {{Defof|Real Area Hyperbolic Cosecant}}
}}
{{eqn | r = \map \ln {\frac a x + \frac {a \paren {\sqrt {\dfrac {a^2 + x^2} {a^2} } } } {\size x} }
      | c = 
}}
{{eqn | r = \map \ln {\frac a x + \frac {a \paren {\dfrac {\sqrt {a^2 + x^2} } a} } {\size x} }
      | c = 
}}
{{eqn | r = \map \ln {\frac a x + \frac {\sqrt {a^2 + x^2} } {\size x} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared plus a squared/Logarithm Form

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \sqrt {x^2 + a^2} \rd x
      | r = \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \arsinh \frac x a + C
      | c = Primitive of $\sqrt {x^2 + a^2}$ in $\arsinh$ form
}}
{{eqn | r = \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \paren {\map \ln {x + \sqrt {x^2 + a^2} } - \ln a} + C
      | c = $\arsinh \dfrac x a$ in Logarithm Form
}}
{{eqn | r = \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } - \frac {a^2 \ln a} 2 + C
      | c = simplifying
}}
{{eqn | r = \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } + C
      | c = subsuming $\dfrac {-a^2 \ln a} 2$ into the arbitrary constant
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared plus a squared/Inverse Hyperbolic Sine Form

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = a \sinh \theta
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \cosh \theta
      | c = Derivative of Hyperbolic Sine
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | l = x
      | r = a \sinh \theta
}}
{{eqn | ll= \leadsto
      | l = x^2 + a^2
      | r = a^2 \sinh^2 \theta + a^2
      | c = 
}}
{{eqn | r = a^2 \paren {\sinh^2 \theta + 1}
      | c = 
}}
{{eqn | r = a^2 \cosh^2 \theta
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \sqrt {x^2 + a^2}
      | r = a \cosh \theta
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = x
      | r = a \sinh \theta
}}
{{eqn | n = 3
      | ll= \leadsto
      | l = \theta
      | r = \sinh^{-1} \frac x a
      | c = {{Defof|Inverse Hyperbolic Sine/Real|index = 1|Real Inverse Hyperbolic Sine}}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \sqrt {x^2 + a^2} \rd x
      | r = \int \sqrt {x^2 + a^2} \, a \cosh \theta \rd \theta
      | c = Integration by Substitution from $(1)$
}}
{{eqn | r = \int a^2 \cosh^2 \theta \rd \theta
      | c = substituting for $\sqrt {x^2 + a^2}$ from $(2)$
}}
{{eqn | r = a^2 \int \cosh^2 \theta \rd \theta
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = a^2 \frac {\sinh \theta \cosh \theta + \theta} 2 + C
      | c = {{Corollary|Primitive of Square of Hyperbolic Cosine Function}}
}}
{{eqn | r = \frac 1 2 a \sinh \theta a \cosh \theta + \frac {a^2 \theta} 2 + C
      | c = rearranging
}}
{{eqn | r = \frac 1 2 x a \cosh \theta + \frac {a^2 \theta} 2 + C
      | c = substituting $x = a \sinh \theta$
}}
{{eqn | r = \frac 1 2 x \sqrt {x^2 + a^2} + \frac {a^2 \theta} 2 + C
      | c = substituting $\sqrt {x^2 + a^2} = a \cosh \theta$ from $(2)$
}}
{{eqn | r = \frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \sinh^{-1} \frac x a + C
      | c = substituting for $\theta$ from $(3)$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over Root of x squared plus a squared/Inverse Hyperbolic Sine Form

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

Proof:
With a view to expressing the problem 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 = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 1
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac x {\sqrt {x^2 + a^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sqrt {x^2 + a^2}
      | c = Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\sqrt {x^2 + a^2} }
      | r = \int x \frac {x \rd x} {\sqrt {x^2 + a^2} }
      | c = 
}}
{{eqn | r = x \sqrt {x^2 + a^2} - \int \sqrt {x^2 + a^2} \rd x
      | c = Integration by Parts
}}
{{eqn | r = x \sqrt {x^2 + a^2} - \left({\frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \sinh^{-1} \frac x a}\right) + C
      | c = Primitive of $\sqrt {x^2 + a^2}$
}}
{{eqn | r = \frac {x \sqrt {x^2 + a^2} } 2 - \frac {a^2} 2 \sinh^{-1} \frac x a + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over Root of x squared plus a squared/Logarithm Form

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

Proof:
With a view to expressing the problem 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 = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 1
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac x {\sqrt {x^2 + a^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sqrt {x^2 + a^2}
      | c = Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\sqrt {x^2 + a^2} }
      | r = \int x \frac {x \rd x} {\sqrt {x^2 + a^2} }
      | c = 
}}
{{eqn | r = x \sqrt {x^2 + a^2} - \int \sqrt {x^2 + a^2} \rd x
      | c = Integration by Parts
}}
{{eqn | r = x \sqrt {x^2 + a^2} - \paren {\frac {x \sqrt {x^2 + a^2} } 2 + \frac {a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } } + C
      | c = Primitive of $\sqrt {x^2 + a^2}$
}}
{{eqn | r = \frac {x \sqrt {x^2 + a^2} } 2 - \frac {a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over Root of x squared minus a squared/Logarithm Form

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

Proof:
With a view to expressing the problem 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 = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 1
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac x {\sqrt {x^2 - a^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sqrt {x^2 - a^2}
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} }
      | r = \int x \frac {x \rd x} {\sqrt {x^2 - a^2} }
      | c = 
}}
{{eqn | r = x \sqrt {x^2 - a^2} - \int \sqrt {x^2 - a^2} \rd x
      | c = Integration by Parts
}}
{{eqn | r = x \sqrt {x^2 - a^2} - \paren {\frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } } + C
      | c = Primitive of $\sqrt {x^2 - a^2}$
}}
{{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \sinh \theta
      | c = Derivative of Hyperbolic Cosine
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | ll= \leadsto
      | l = x^2 - a^2
      | r = a^2 \cosh^2 \theta - a^2
      | c = 
}}
{{eqn | r = a^2 \paren {\cosh^2 \theta - 1}
      | c = 
}}
{{eqn | r = a^2 \sinh^2 \theta
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \sqrt {x^2 - a^2}
      | r = a \sinh \theta
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | n = 3
      | ll= \leadsto
      | l = \theta
      | r = \cosh^{-1} \frac x a
      | c = {{Defof|Inverse Hyperbolic Cosine/Real|index = 1|Real Inverse Hyperbolic Cosine}}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \sqrt {x^2 - a^2} \rd x
      | r = \int \sqrt {x^2 - a^2} \, a \sinh \theta \rd \theta
      | c = Integration by Substitution from $(1)$
}}
{{eqn | r = \int a^2 \sinh^2 \theta \rd \theta
      | c = substituting for $\sqrt {x^2 - a^2}$ from $(2)$
}}
{{eqn | r = a^2 \int \sinh^2 \theta \rd \theta
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = a^2 \frac {\sinh \theta \cosh \theta - \theta} 2 + C
      | c = Primitive of $\sinh^2 \theta$: Corollary
}}
{{eqn | r = \frac 1 2 a \sinh \theta a \cosh \theta - \frac {a^2 \theta} 2 + C
      | c = rearranging
}}
{{eqn | r = \frac 1 2 x a \sinh \theta - \frac {a^2 \theta} 2 + C
      | c = substituting $x = a \cosh \theta$
}}
{{eqn | r = \frac 1 2 x \sqrt {x^2 - a^2} - \frac {a^2 \theta} 2 + C
      | c = substituting $\sqrt {x^2 - a^2} = a \sinh \theta$ from $(2)$
}}
{{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \cosh^{-1} \frac x a + C
      | c = substituting for $\theta$ from $(3)$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over Root of x squared minus a squared/Inverse Hyperbolic Cosine Form

:$\ds \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } = \frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \cosh^{-1} \frac x a + C$
for $x > a$.

Proof:
With a view to expressing the problem 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 = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 1
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac x {\sqrt {x^2 - a^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sqrt {x^2 - a^2}
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} }
      | r = \int x \frac {x \rd x} {\sqrt {x^2 - a^2} }
      | c = 
}}
{{eqn | r = x \sqrt {x^2 - a^2} - \int \sqrt {x^2 - a^2} \rd x
      | c = Integration by Parts
}}
{{eqn | r = x \sqrt {x^2 - a^2} - \paren {\frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \cosh^{-1} \frac x a} + C
      | c = Primitive of $\sqrt {x^2 - a^2}$
}}
{{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \cosh^{-1} \frac x a + C
      | c = simplifying
}}
{{end-eqn}}
Note that because:
:$\cosh^{-1} \dfrac x a$ is defined for $x \ge a$ only
and:
:$\dfrac {x^2} {\sqrt {x^2 - a^2} }$ is not defined for $x = a$
$x$ is constrained as indicated.
{{qed}}
Theorem: Primitive of x by Inverse Hyperbolic Cosine of x over a

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

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
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \arcosh \frac x a \rd x
      | r = \frac {x^2} 2 \arcosh \frac x a - \int \frac {x^2} 2 \paren {\frac 1 {\sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^2} 2 \arcosh \frac x a - \frac 1 2 \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {x^2} 2 \arcosh \frac x a - \frac 1 2 \paren {\frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \arcosh \frac x a} + C
      | c = Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \paren {\frac {x^2} 2 - \frac {a^2} 4} \arcosh \frac x a - \frac {x \sqrt {x^2 - a^2} } 4 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x squared minus a squared/Logarithm Form

:$\ds \int \frac {x^2 \rd x} {x^2 - a^2} = x + \frac a 2 \map \ln {\frac {x - a} {x + a} } + C$
for $x^2 > a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {x^2 - a^2}
      | r = \int \frac {x^2 - a^2 + a^2} {x^2 - a^2} \rd x
      | c = 
}}
{{eqn | r = \int \frac {x^2 - a^2} {x^2 - a^2} \rd x + \int \frac {a^2} {x^2 - a^2} \rd x
      | c = 
}}
{{eqn | r = \int \d x + a^2 \int \frac {\d x} {x^2 - a^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x + a^2 \int \frac {\d x} {x^2 - a^2} + C
      | c = Primitive of Constant
}}
{{eqn | r = x + a^2 \paren {\frac 1 {2 a} \map \ln {\frac {x - a} {x + a} } } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form
}}
{{eqn | r = x + \frac a 2 \map \ln {\frac {x - a} {x + a} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x squared minus a squared/Inverse Hyperbolic Cotangent Form

:$\ds \int \frac {x^2 \rd x} {x^2 - a^2} = x - a \coth^{-1} \frac x a + C$
for $x^2 > a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {x^2 - a^2}
      | r = \int \frac {x^2 - a^2 + a^2} {x^2 - a^2} \rd x
      | c = 
}}
{{eqn | r = \int \frac {x^2 - a^2} {x^2 - a^2} \rd x + \int \frac {a^2} {x^2 - a^2} \rd x
      | c = 
}}
{{eqn | r = \int \d x + a^2 \int \frac {\d x} {x^2 - a^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = x + a^2 \int \frac {\d x} {x^2 - a^2} + C
      | c = Primitive of Constant
}}
{{eqn | r = x + a^2 \paren {-\frac 1 a \coth^{-1} {\frac x a} } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$: $\coth^{-1}$ form
}}
{{eqn | r = x - a \coth^{-1} {\frac x a} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Category:Primitive of x squared over x squared minus a squared
ddokvn26vs5e79f0sk1wwvvgwgo9w92
Theorem: Primitive of x squared over a squared minus x squared/Logarithm Form

:$\ds \int \frac {x^2 \rd x} {a^2 - x^2} = -x + \frac a 2 \map \ln {\frac {a + x} {a - x} } + C$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {a^2 - x^2}
      | r = \int \frac {x^2 - a^2 + a^2} {a^2 - x^2} \rd x
      | c = 
}}
{{eqn | r = \int \frac {-\paren {a^2 - x^2} } {a^2 - x^2} \rd x + \int \frac {a^2} {a^2 - x^2} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = -\int \d x + a^2 \int \frac {\d x} {a^2 - x^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = -x + a^2 \int \frac {\d x} {a^2 - x^2} + C
      | c = Primitive of Constant
}}
{{eqn | r = -x + a^2 \paren {\frac 1 {2 a} \map \ln {\frac {a + x} {a - x} } } + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$: Logarithm Form
}}
{{eqn | r = -x + \frac a 2 \map \ln {\frac {a + x} {a - x} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over a squared minus x squared/Inverse Hyperbolic Tangent Form

:$\ds \int \frac {x^2 \rd x} {a^2 - x^2} = -x + a \tanh^{-1} \frac x a + C$
for $x^2 < a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {a^2 - x^2}
      | r = \int \frac {x^2 - a^2 + a^2} {a^2 - x^2} \rd x
      | c = 
}}
{{eqn | r = \int \frac {-\paren {a^2 - x^2} } {a^2 - x^2} \rd x + \int \frac {a^2} {a^2 - x^2} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = -\int \rd x + a^2 \int \frac {\rd x} {a^2 - x^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = -x + a^2 \int \frac {\rd x} {a^2 - x^2} + C
      | c = Primitive of Constant
}}
{{eqn | r = -x + a^2 \paren {\frac 1 a \tanh^{-1} \frac x a} + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$: $\tanh^{-1}$ form
}}
{{eqn | r = -x + a \tanh^{-1} \frac x a + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Category:Primitive of x squared over a squared minus x squared
5we4g2qndadl4yqvbmxdbgzj4rrrwv4
Theorem: Primitive of x by Inverse Hyperbolic Tangent of x over a

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

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 = \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
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^2} 2
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \artanh \frac x a \rd x
      | r = \frac {x^2} 2 \artanh \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 \artanh \frac x a - \frac a 2 \int \frac {x^2 \rd x} {a^2 - x^2} + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^2} 2 \artanh \frac x a - \frac a 2 \paren {-x + a \artanh \frac x a} + C
      | c = Primitive of $\dfrac {x^2} {a^2 - x^2}$
}}
{{eqn | r = \frac {a x} 2 + \frac {x^2 - a^2} 2 \artanh \frac x a + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}