avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of x squared by Root of x squared plus a squared

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

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 \sqrt {x^2 + a^2} \rd x
      | r = \int \frac {z \sqrt {z + a^2} \rd z} {2 \sqrt z}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \sqrt z \sqrt {z + a^2} \rd z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {\frac {2 z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C
      | c = Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$
}}
{{eqn | r = \frac {2 z + a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} {16} \paren {2 \map \ln {\sqrt {z + a^2} + \sqrt z} } + C
      | c = Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$
}}
{{eqn | r = \frac {2 z + 2 a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C
      | c = 
}}
{{eqn | r = \frac {z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C
      | c = 
}}
{{eqn | r = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 - \frac {a^2 x \sqrt {x^2 + a^2} } 8 - \frac {a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed by Root of x squared plus 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 plus a squared over x cubed

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

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 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} } + 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} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Root of x squared plus 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:
{{begin-eqn}}
{{eqn | l = x
      | r = a \tan \theta
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \sec^2 \theta
      | c = Derivative of Tangent Function
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}
      | r = \int \frac {a \sec^2 \theta \rd \theta} {\sqrt {a^2 \tan^2 \theta + a^2}^3}
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac {a \sec^2 \theta \rd \theta} {a^3 \sec^3 \theta}
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \dfrac 1 {a^2} \int \cos \theta \rd \theta
      | c = {{Defof|Real Secant Function}} and simplification
}}
{{eqn | r = \dfrac 1 {a^2} \sin \theta + C
      | c = Primitive of $\cos x$
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \sin \theta} {\cos \theta} \dfrac {\cos \theta} a + C
      | c = 
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \tan \theta} {a \sec \theta} + C
      | c = Tangent is Sine divided by Cosine, {{Defof|Real Secant Function}}
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \tan \theta} {\sqrt {a^2 \sec^2 \theta} } + C
      | c = 
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \tan \theta} {\sqrt {a^2 \tan^2 \theta + a^2} } + C
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac x {\sqrt {x^2 + a^2} } + C
      | c = substituting $x = a \tan \theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Root of x squared plus 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 plus a squared cubed

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

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
}}
{{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} } + \map \ln {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 plus 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 plus a squared cubed

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

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 {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} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + 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 plus 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 plus a squared cubed

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

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} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} } - \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} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac 1 {a^2} \paren {\frac 1 {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} } + 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} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared plus a squared cubed

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

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 z} {\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 \map \ln {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 plus 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 plus a squared cubed

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

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 \map \ln {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} \map \ln {x + \sqrt {x^2 + a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed by Root of x squared plus a squared cubed

:$\ds \int x^3 \paren {\sqrt {x^2 + a^2} }^3 \rd x = \frac {\paren {\sqrt {x^2 + a^2} }^7} 7 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^5} 5 + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int x^3 \paren {\sqrt {x^2 + a^2} }^3 \rd x
      | r = \int x \paren {x^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x
      | c = 
}}
{{eqn | r = \int x \paren {x^2 + a^2 - a^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x
      | c = Primitive of Power
}}
{{eqn | r = \int x \paren {x^2 + a^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int x \paren {\sqrt {x^2 + a^2} }^5 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x
      | c = simplifying
}}
{{end-eqn}}
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
      | o = 
      | r = \int x \paren {\sqrt {x^2 + a^2} }^5 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x
      | c = 
}}
{{eqn | r = \int \frac {\sqrt z z^{5/2} } {2 \sqrt z} \rd z - a^2 \int \frac {\sqrt z z^{3/2} } {2 \sqrt z} \rd z
      | c = Integration by Substitution
}}
{{eqn | r =  \frac 1 2 \int z^{5/2} \rd z - \frac {a^2} 2 \int z^{3/2} \rd z
      | c = simplifying
}}
{{eqn | r = \frac 1 2 \frac {z^{7/2} } {7/2} - \frac {a^2} 2 \frac {z^{5/2} } {5/2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {z^{7/2} } 7 - a^2 \frac {z^{5/2} } 5 + C
      | c = simplifying
}}
{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^7} 7 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^5} 5 + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared plus a squared cubed over x

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

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 {\paren {\sqrt {x^2 + a^2} }^3} x \rd x
      | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z} \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \int \frac {\sqrt {z + a^2} } z \rd z
      | c = Primitive of $\dfrac {\paren {\sqrt {a x + b} }^m} x$
}}
{{eqn | r = \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \paren {2 \sqrt {z + a^2} + a^2 \int \frac {\d z} {z \sqrt {z + a^2} } }
      | c = Primitive of $\dfrac {\sqrt {a x + b} } x$
}}
{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} + a^4 \int \frac {\d x} {x \sqrt {x^2 + a^2} }
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} - a^3 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
      | c = Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared plus a squared cubed over x squared

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

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 \frac {\paren {\sqrt {x^2 + a^2} }^3} {x^2} \rd x
      | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z \sqrt z} \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac {-\paren {\sqrt {z + a^2} }^3} {\sqrt z} + \frac 3 2 \int \frac {\sqrt {z + a^2} } {\sqrt z} \rd z
      | c = Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3
}}
{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} x + 3 \int \sqrt {x^2 + a^2} \rd x
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} x + \frac {3 x \sqrt {x^2 + a^2} } 2 + \frac {3 a^2} 2 \map \ln {x + \sqrt {x^2 + a^2} } + C
      | c = Primitive of $\sqrt {x^2 + a^2}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of x squared plus a squared cubed over x cubed

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

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 {\paren {\sqrt {x^2 + a^2} }^3} {x^3} \rd x
      | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z^2} \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \paren {\frac {-\paren {\sqrt {z + a^2} }^3} z + \frac 3 2 \int \frac {\sqrt {z + a^2} } z \rd z}
      | c = Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3
}}
{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \int \frac {\sqrt {x^2 + a^2} } x \rd x
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \paren {\sqrt {x^2 + a^2} - a \map \ln {\frac {a + \sqrt {x^2 + a^2} } a} } + C
      | c = Primitive of $\dfrac {\sqrt {x^2 + a^2} } x$
}}
{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 + a^2} } 2 - \frac {3 a} 2 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}