avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of x over Root of x squared minus a squared

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z^2
      | r = x^2 - a^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 z \frac {\d z} {\d x}
      | r = 2 x
      | c = Chain Rule for Derivatives, Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {\sqrt {x^2 - a^2} }
      | r = \int \frac {z \rd z} z
      | c = Integration by Substitution
}}
{{eqn | r = \int \rd z
      | c = 
}}
{{eqn | r = z + C
      | c = Primitive of Constant
}}
{{eqn | r = \sqrt {x^2 - a^2} + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over Root of x squared minus a squared

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

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^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 2 x
      | 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^3 \rd x} {\sqrt {x^2 - a^2} }
      | r = \int x^2 \frac {x \rd x} {\sqrt {x^2 - a^2} }
      | c = 
}}
{{eqn | r = x^2 \sqrt {x^2 - a^2} - \int 2 x \sqrt {x^2 - a^2} \rd x
      | c = Integration by Parts
}}
{{eqn | r = x^2 \sqrt {x^2 - a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 - a^2} }^3} 3} + C
      | c = Primitive of $x \sqrt {x^2 - a^2}$
}}
{{eqn | r = \paren {x^2 - a^2 + a^2} \sqrt {x^2 - a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 - a^2} }^3} 3} + C
      | c = 
}}
{{eqn | r = \paren {\sqrt {x^2 - a^2} }^3 + a^2 \sqrt {x^2 - a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 - a^2} }^3} 3} + C
      | c = 
}}
{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 + a^2 \sqrt {x^2 - a^2} + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by Root of x squared minus a squared

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \sinh \theta
      | c = Derivative of Hyperbolic Cosine
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | ll= \leadsto
      | l = \sqrt {x^2 - a^2}
      | r = \sqrt {a^2 \paren {\cosh^2 \theta - 1} }
      | c = 
}}
{{eqn | r = \sqrt {a^2 \sinh^2 \theta}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = a \sinh \theta
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }
      | r = \int \frac {a \sinh \theta \rd \theta} {a^2 \cosh^2 \theta \cdot a \sinh \theta}
      | c = Integration by Substitution
}}
{{eqn | r = \dfrac 1 {a^2} \int \frac {\rd \theta} {\cosh^2 \theta}
      | c = simplifying
}}
{{eqn | r = \dfrac 1 {a^2} \tanh \theta + C
      | c = Primitive of $\dfrac 1 {\cosh^2 \theta}$
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \sinh \theta} {a \cosh \theta} + C
      | c = {{Defof|Hyperbolic Tangent|index = 2}}
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {\sqrt {x^2 - a^2} } x + C
      | c = substituting for $a \sinh \theta$ and $a \cosh \theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by Root of x squared minus a squared

:$\ds \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} }
      | r = \int \frac {\d z} {2 z^{3/2} \sqrt z \sqrt {z - a^2} }
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\d z} {z^2 \sqrt {z - a^2} }
      | c = 
}}
{{end-eqn}}
Using Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$:
:$\ds \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m-1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$
Setting:
{{begin-eqn}}
{{eqn | l = x
      | o = :=
      | r = z
}}
{{eqn | l = m
      | o = :=
      | r = 2
}}
{{eqn | l = a
      | o = :=
      | r = 1
}}
{{eqn | l = b
      | o = :=
      | r = -a^2
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \frac 1 2 \int \frac {\d z} {z^2 \sqrt {z - a^2} }
      | r = \frac {-\sqrt {z - a^2} } {2 \paren {2 - 1} \paren {-a^2} z^{2 - 1} } - \frac {2 \paren 2 - 3} {2 \paren {2 \paren 2 - 2} \paren {-a^2} } \int \frac {\d z} {z^{2 - 1} \sqrt {z - a^2} } + C
      | c = Primitive of $ \dfrac 1 {x^m \sqrt{a x + b} }$
}}
{{eqn | r = \frac {\sqrt {z - a^2} } {2 a^2 z} + \frac 1 {4 a^2} \int \frac {\d z} {z \sqrt{z - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {4 a^2} \paren {\int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } } + C
      | c = substituting for $z$
}}
{{eqn | r = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^2} \paren {\int \frac {\d x} {x \sqrt {x^2 - a^2} } } + C
      | c = simplification
}}
{{eqn | r = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Root of x squared minus a squared

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int x \sqrt {x^2 - a^2} \rd x
      | r = \int \frac {\sqrt z \sqrt {z - a^2} \rd z} {2 \sqrt z}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \sqrt {z - a^2} \rd z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \frac {2 \paren {\sqrt {z - a^2} }^3} 3 + C
      | c = Primitive of $\sqrt {a x + b}$
}}
{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 + C
      | c = substituting for $z$ and simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed by Root of x squared minus a squared

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int x^3 \sqrt {x^2 - a^2} \rd x
      | r = \int \frac {z^{3/2} \sqrt {z - a^2} \rd z} {2 \sqrt z}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int z \sqrt {z - a^2} \rd z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {\frac {2 \paren {3 z + 2 a^2} } {15} \paren {\sqrt {z - a^2} }^3 } + C
      | c = Primitive of $x \sqrt {\paren {a x + b} }$
}}
{{eqn | r = \frac {3 z + 2 a^2} {15} \paren {\sqrt {z - a^2} }^3 + C
      | c = simplification
}}
{{eqn | r = \frac {3 x^2 + 2 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C
      | c = substituting for $z$
}}
{{eqn | r = \frac {3 x^2 - 3 a^2 + 5 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C
      | c = 
}}
{{eqn | r = \frac {3 \paren {x^2 - a^2} \paren {\sqrt {a^2 - x^2} }^3 + 5 a^2 \paren {\sqrt {a^2 - x^2} }^3} {15} + C
      | c = 
}}
{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^3} 3 + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared minus a squared over x

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\sqrt {x^2 - a^2} } x \rd x
      | r = \int \frac {\sqrt {z - a^2} \rd z} {2 \sqrt z \sqrt z}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\sqrt {z - a^2} \rd z} z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {2 \sqrt {z - a^2} - a^2 \int \frac {\d z} {z \sqrt {z - a^2} } } + C
      | c = Primitive of $\dfrac {\sqrt {a x + b} } x$
}}
{{eqn | r = \sqrt {x^2 - a^2} - \frac {a^2} 2 \int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } + C
      | c = substituting for $z$
}}
{{eqn | r = \sqrt {x^2 - a^2} - a^2 \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \sqrt {x^2 - a^2} - a^2 \paren {\frac 1 a \arcsec \size {\frac x a} } + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$
}}
{{eqn | r = \sqrt {x^2 - a^2} - a \arcsec \size {\frac x a} + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared minus a squared over x cubed

:$\ds \int \frac {\sqrt {x^2 - a^2} } {x^3} \rd x = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 {2 a} \arcsec \size {\frac x a} + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\sqrt {x^2 - a^2} } {x^3} \rd x
      | r = \int \frac {\sqrt {z - a^2} \rd z} {2 z^{3/2} \sqrt z}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\sqrt {z - a^2} \rd z} {z^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {\frac {-\sqrt {z - a^2} } z + \frac 1 2 \int \frac {\d z} {z \sqrt {z - a^2} } } + C
      | c = Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 4 \int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } + C
      | c = substituting for $z$
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 2 \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 2 \paren {\frac 1 a \arcsec \size {\frac x a} } + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 {2 a} \arcsec \size {\frac x a} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Root of x squared minus a squared cubed

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3}
      | r = \int \frac {\d z} {2 \sqrt z \paren {\sqrt {z - a^2} }^3}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\d z} {\paren {z - a^2} \sqrt z \sqrt {z - a^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {\frac {2 \sqrt z} {\paren {-a^2} \sqrt {z - a^2} } } + C
      | c = Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$
}}
{{eqn | r = \frac 1 2 \paren {\frac {2 x} {\paren {-a^2} \sqrt {x^2 - a^2} } } + C
      | c = substituting for $z$
}}
{{eqn | r = \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Root of x squared minus a squared cubed

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2 - a^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | r = \int \frac {x \rd z} {2 x z^{3/2} }
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int z^{-3/2} \rd z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {\frac {-z^{1/2} } {\frac 1 2} } + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {\sqrt z} + C
      | c = simplifying
}}
{{eqn | r = \frac {-1} {\sqrt {x^2 - a^2} } + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over Root of x squared minus a squared cubed

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

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
}}
{{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 {\paren {\sqrt {x^2 - a^2} }^3}
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} {\sqrt {x^2 - a^2} }
      | c = Primitive of $\dfrac x {\paren {\sqrt {x^2 - a^2} }^3}$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | r = \int x \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | c = 
}}
{{eqn | r = x \frac {-1} {\sqrt {x^2 - a^2} } - \int \frac {-1} {\sqrt {x^2 - a^2} } 1 + C
      | c = Integration by Parts
}}
{{eqn | r = \frac {-x} {\sqrt {x^2 - a^2} } + \int \frac 1 {\sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {-x} {\sqrt {x^2 - a^2} } + \ln \size {x + \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over Root of x squared minus a squared cubed

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^3 \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | r = \int \frac {x \paren {x^2} \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | c = 
}}
{{eqn | r = \int \frac {x \paren {x^2 - a^2 + a^2} \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | c = 
}}
{{eqn | r = \int \frac {x \paren {x^2 - a^2} \rd x} {\paren {\sqrt {x^2 - a^2} }^3} + a^2 \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + a^2 \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
      | c = simplifying
}}
{{eqn | r = \sqrt {x^2 - a^2} + a^2 \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3} + C
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{eqn | r = \sqrt {x^2 - a^2} + a^2 \frac {-1} {\sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac x {\paren {\sqrt {x^2 - a^2} }^3}$
}}
{{eqn | r = \sqrt {x^2 - a^2} - \frac {a^2} {\sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Root of x squared minus a squared cubed

:$\ds \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3} = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3}
      | r = \int \frac {\d z} {2 \sqrt z \sqrt z \paren {\sqrt {z - a^2} }^3}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\d z} {z \paren {\sqrt {z - a^2} }^3}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {\frac 2 {\paren {-a^2} \sqrt {z - a^2} } + \frac 1 {-a^2} \int \frac {\d z} {z \sqrt {z - a^2} } } + C
      | c = Primitive of $\dfrac 1 {x \paren {\sqrt{a x + b} }^m}$
}}
{{eqn | r = \frac {-1} {a^2 \sqrt {z - a^2} } - \frac 1 {2 a^2} \int \frac {\d z} {z \sqrt {z - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } + C
      | c = substituting for $z$
}}
{{eqn | r = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^2} \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by Root of x squared minus a squared cubed

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | r = \int \frac {a^2 \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | c = 
}}
{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | c = 
}}
{{eqn | r = \frac 1 {a^2} \int \frac {-\paren {x^2 - a^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3} + \frac 1 {a^2} \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {-1} {a^2} \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} } + \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3}
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^2} \frac {\sqrt {x^2 - a^2} } {a^2 x} + \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3} + C
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {a^4 x} + \frac 1 {a^2} \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac 1 {\paren {\sqrt {x^2 - a^2} }^3}$
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 - a^2} } + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by Root of x squared minus a squared cubed

:$\ds \int \frac {\d x} {x^3 \paren {\sqrt {x^2 - a^2} }^3} = \frac 1 {2 a^2 x^2 \sqrt {x^2 - a^2} } - \frac 3 {2 a^4 \sqrt {x^2 - a^2} } + \frac 3 {2 a^5} \arcsec \size {\frac x a} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {\sqrt {x^2 - a^2} }^3}
      | r = \int \frac {a^2 \rd x} {a^2 x^3 \paren {\sqrt {x^2 - a^2} }^3}
      | c = 
}}
{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^3 \paren {\sqrt {x^2 - a^2} }^3}
      | c = 
}}
{{eqn | r = \frac 1 {a^2} \int \frac {-\paren {x^2 - a^2} \rd x} {x^3 \paren {\sqrt {x^2 - a^2} }^3} + \frac 1 {a^2} \int \frac {x^2 \rd x} {x^3 \paren {\sqrt {x^2 - a^2} }^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {-1} {a^2} \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3}
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^2} \paren {\frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3} + C
      | c = Primitive of $\dfrac 1 {x^3 \sqrt {x^2 - a^2} }$
}}
{{eqn | r = \frac {-1} {a^2} \paren {\frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} }
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \frac 1 {a^2} \paren {\frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} } + C
      | c = Primitive of $\dfrac 1 {x \paren {\sqrt {x^2 - a^2} }^3}$
}}
{{eqn | r = \frac 1 {2 a^2 x^2 \sqrt {x^2 - a^2} } - \frac 3 {2 a^4 \sqrt {x^2 - a^2} } + \frac 3 {2 a^5} \arcsec \size {\frac x a} + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared minus a squared cubed

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \paren {\sqrt {x^2 - a^2} }^3 \rd x
      | r = \int \frac {\paren {\sqrt {z - a^2} }^3} {2 \sqrt z} \rd x
      | c = Integration by Substitution
}}
{{eqn | r = \frac {\sqrt z \paren {\sqrt {z - a^2} }^3} 4 - \frac {3 a^2} 8 \int \frac {\sqrt {z - a^2} } {\sqrt z} \rd x + C
      | c = Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt{a x + b} }$
}}
{{eqn | r = \frac {\sqrt z \paren {\sqrt {z - a^2} }^3} 4 - \frac {3 a^2} 8 \paren {\sqrt z \sqrt {z - a^2} - \frac {a^2} 2 \int \frac {\d x} {\sqrt z \sqrt {z - a^2} } } + C
      | c = Primitive of $\dfrac {\sqrt {p x + q} } {\sqrt{a x + b} }$
}}
{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2} 8 x \sqrt {x^2 + a^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {x^2 - a^2} } + C
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Root of x squared minus a squared cubed

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2 - a^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x
      | r = \int \frac {z^{3/2} } 2 \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \frac {z^{5/2} } {\frac 5 2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {z^{5/2} } 5 + C
      | c = simplifying
}}
{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Root of x squared minus a squared cubed

:$\ds \int x^2 \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^5} 6 + \frac {a^2 x \paren {\sqrt {x^2 - a^2} }^3} {24} - \frac {a^4 x \sqrt {x^2 - a^2} } {16} + \frac {a^6} {16} \ln \size {x + \sqrt {x^2 - a^2} } + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int x^2 \paren {\sqrt {x^2 - a^2} }^3 \rd x
      | r = \int \frac {\sqrt z \paren {z - a^2}^{3/2} } 2 \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac {\sqrt z \paren {z - a^2}^{5/2} } 6 + \frac {a^2} {12} \int \frac {\paren {z - a^2}^{3/2} } {\sqrt z} \rd z + C
      | c = Primitive of $\paren {p x + q}^n \sqrt{a x + b}$
}}
{{eqn | r = \frac {x \paren {x^2 - a^2}^{5/2} } 6 + \frac {a^2} 6 \int \paren {x^2 - a^2}^{3/2} \rd x + C
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {x \paren {x^2 - a^2}^{5/2} } 6
}}
{{eqn | o =
      | ro= +
      | r = \frac {a^2} 6 \paren {\frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } } + C
      | c = Primitive of $\paren {\sqrt {x^2 - a^2} }^3$
}}
{{eqn | r = \frac {x \paren {x^2 - a^2}^{5/2} } 6 + \frac {a^2 x \paren {\sqrt {x^2 - a^2} }^3} {24}
}}
{{eqn | o =
      | ro= -
      | r = \frac {a^4 x \sqrt {x^2 - a^2} } {16} + \frac {a^6} {16} \ln \size {x + \sqrt {x^2 - a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}