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

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

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

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

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

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

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 {a^2 - x^2} }^3}
      | r = \int \frac {\d z} {2 \sqrt z \sqrt z \paren {\sqrt {a^2 - z} }^3}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\d z} {z \paren {\sqrt {a^2 - z} }^3}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \paren {\frac 2 {a^2 \sqrt {a^2 - z} } + \frac 1 {a^2} \int \frac {\d z} {z \sqrt {a^2 - z} } } + C
      | c = Primitive of $\dfrac 1 {x \paren {\sqrt {a x + b} }^m }$
}}
{{eqn | r = \frac 1 {a^2 \sqrt {a^2 - z} } + \frac 1 {2 a^2} \int \frac {\d z} {z \sqrt {a^2 - z} } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^2 \sqrt {a^2 - x^2} } + \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \sqrt {a^2 - x^2} } + C
      | c = substituting for $z$
}}
{{eqn | r = \frac 1 {a^2 \sqrt {a^2 - x^2} } + \frac 1 {a^2} \int \frac {\d x} {x \sqrt {a^2 - x^2} } + C
      | c = simplifying
}}
{{eqn | r = \frac1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C
      | c = Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by Root of a squared minus x squared cubed

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

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

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

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

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

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 \paren {\sqrt {a^2 - x^2} }^3 \rd x
      | r = \int \frac {\paren {\sqrt {a^2 - z} }^3} {2 \sqrt z} \rd x
      | c = Integration by Substitution
}}
{{eqn | r = \frac {\sqrt z \paren {\sqrt {a^2 - z} }^3} 4 + \frac {3 a^2} 8 \int \frac {\sqrt {a^2 - z} } {\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 {a^2 - z} }^3} 4 + \frac {3 a^2} 8 \paren {\sqrt z \sqrt {a^2 - z} + \frac {a^2} 2 \int \frac {\d x} {\sqrt z \sqrt {a^2 - z} } } + C
      | c = Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$
}}
{{eqn | r = \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2} 8 x \sqrt {a^2 - x^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {a^2 - x^2} } + C
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a + C
      | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Root of a squared minus x squared cubed

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a^2 - 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 \paren {\sqrt {a^2 - x^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 {a^2 - x^2} }^5} 5 + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Root of a squared minus x squared cubed

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

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 x^2 \paren {\sqrt {a^2 - x^2} }^3 \rd x
      | r = \int \frac {\sqrt z \paren {a^2 - z}^{3/2} } 2 \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac {\sqrt z \paren {a^2 - z}^{5/2} } {-6} + \frac {-a^2} {-12} \int \frac {\paren {a^2 - z}^{3/2} } {\sqrt z} \rd z + C
      | c = Primitive of $\paren {p x + q}^n \sqrt{a x + b}$
}}
{{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2} 6 \int \paren {a^2 - x^2}^{3/2} \rd x + C
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6
}}
{{eqn | o =
      | ro= +
      | r = \frac {a^2} 6 \paren {\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a} + C
      | c = Primitive of $\paren {\sqrt {a^2 - x^2} }^3$
}}
{{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2 x \paren {\sqrt {a^2 - x^2} }^3} {24}
}}
{{eqn | o =
      | ro= +
      | r = \frac {a^4 x \sqrt {a^2 - x^2} } {16} + \frac {a^6} {16} \arcsin \frac x a + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed by Root of a squared minus x squared cubed

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

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

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

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

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

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 {\paren {\sqrt {a^2 - x^2} }^3} {x^2} \rd x
      | r = \int \frac {\paren {\sqrt {a^2 - z} }^3} {2 z \sqrt z} \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac {-\paren {\sqrt {a^2 - z} }^3} {\sqrt z} - \frac 3 2 \int \frac {\sqrt {a^2 - z} } {\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 {a^2 - x^2} }^3} x - 3 \int \sqrt {a^2 - x^2} \rd x
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {-\paren {\sqrt {a^2 - x^2} }^3} x - \frac {3 x \sqrt {a^2 - x^2} } 2 - \frac {3 a^2} 2 \arcsin \frac x a + C
      | c = Primitive of $\sqrt {a^2 - x^2}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a squared minus x squared cubed over x cubed

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

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

:<nowiki>
$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end {cases}$
</nowiki>

Proof:
First:
{{begin-eqn}}
{{eqn | l = a x^2 + b x + c
      | r = \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}
      | c = Completing the Square
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }
      | c = 
}}
{{end-eqn}}
Put:
{{begin-eqn}}
{{eqn | l = z
      | r = 2 a x + b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a
      | c = Derivative of Power
}}
{{end-eqn}}
Let $D = b^2 - 4 a c$.
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }
      | c = from $(1)$
}}
{{eqn | r = \int \frac {2 \rd z} {z^2 - D}
      | c = Integration by Substitution
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c

:<nowiki>
$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end {cases}$
</nowiki>

Proof:
We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:
{{begin-eqn}}
{{eqn | l = a
      | r = 3
}}
{{eqn | l = b
      | r = 4
}}
{{eqn | l = c
      | r = 2
}}
{{end-eqn}}
We note that:
{{begin-eqn}}
{{eqn | l = b^2 - 4 a c
      | r = 4^2 - 4 \times 3 \times 2
}}
{{eqn | r = 16 - 24
}}
{{eqn | r = -8
}}
{{end-eqn}}
Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
Substituting for $a$, $b$ and $c$ and simplifying:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {3 x^2 + 4 x + 2}
      | r = \dfrac 2 {\sqrt {4 \times 3 \times 2 - 4^2} } \map \arctan {\dfrac {2 \times 3 x + 4} {\sqrt {4 \times 3 \times 2 - 4^2} } } + C
      | c = 
}}
{{eqn | r = \dfrac 2 {\sqrt 8} \map \arctan {\dfrac {6 x + 4} {\sqrt 8} } + C
      | c = 
}}
{{eqn | r = \dfrac 2 {2 \sqrt 2} \map \arctan {\dfrac {2 \paren {3 x + 2} } {2 \sqrt 2} } + C
      | c = 
}}
{{eqn | r = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c

:<nowiki>
$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end {cases}$
</nowiki>

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {3 x^2 + 4 x + 2}
      | r = \dfrac 1 3 \int \frac {\d x} {x^2 + \frac 4 3 x + \frac 2 3}
      | c = 
}}
{{eqn | r = \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \paren {\frac 2 3 - \frac 4 9} }
      | c = 
}}
{{eqn | r = \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \frac 2 9}
      | c = 
}}
{{eqn | r = \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \paren {\frac {\sqrt 2} 3}^2}
      | c = 
}}
{{eqn | r = \dfrac 1 3 \paren {\dfrac 3 {\sqrt 2} \map \arctan {\dfrac {x + \frac 2 3} {\frac {\sqrt 2} 3} } } + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c

:<nowiki>
$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end {cases}$
</nowiki>

Proof:
First:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {\d x} {a x^2 + c}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {\d x} {x^2 + \frac c a}
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Let $a c > 0$.
Then $\dfrac c a > 0$ and:
{{begin-eqn}}
{{eqn | l = \frac c a
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac c a
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {\frac c a}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \frac 1 a \int \frac {\d x} {x^2 + q^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a q} \arctan \frac x q + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \frac 1 {a \sqrt {\frac c a} } \arctan \frac x {\sqrt {\frac c a} } + C
      | c = substituting for $q$
}}
{{eqn | r = \dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $a c < 0$.
Then $\dfrac c a > 0$ and:
{{begin-eqn}}
{{eqn | l = \frac c a
      | o = <
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\frac c a
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {-\frac c a}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \frac 1 a \int \frac {\d x} {x^2 - q^2}
      | c = 
}}
{{eqn | r = \frac 1 {2 a q} \ln \size {\frac {z - q} {z + q} } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {2 a \sqrt {-\frac c a} } \ln \size {\frac {z - \sqrt {-\frac c a} } {z + \sqrt {-\frac c a} } } + C
      | c = substituting for $q$
}}
{{eqn | r = \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $c = 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {\d x} {a x^2}
      | c =
}}
{{eqn | r = \dfrac {-1} {a x} + C
      | c = Primitive of Power
}}
{{end-eqn}}
{{qed}}