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

:$\ds \int \frac {x^2 \rd x} {\paren {a^2 - x^2}^2} = \frac x {2 \paren {a^2 - x^2} } - \frac 1 {4 a} \map \ln {\frac {a + x} {a - x} } + C$
for $x^2 < a^2$.

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

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

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

:$\ds \int \frac {\d x} {x \paren {a^2 - x^2}^2} = \frac 1 {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {a^2 - x^2}^2}
      | r = \int \paren {\frac 1 {a^4 x} + \frac x {a^4 \paren {a^2 - x^2} } + \frac x {a^2 \paren {a^2 - x^2}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} x + \frac 1 {a^4} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \paren {-\frac 1 2 \map \ln {a^2 - x^2} } + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
      | c = Primitive of $\dfrac x {a^2 - x^2}$
}}
{{eqn | r = \frac 1 {a^4} \ln \size x - \frac 1 {2 a^4} \map \ln {a^2 - x^2} + \frac 1 {a^2} \paren {\frac 1 {2 \paren {a^2 - x^2} } } + C
      | c = Primitive of $\dfrac x {\paren {a^2 - x^2}^2}$
}}
{{eqn | r = \frac 1 {2 a^4} \map \ln {x^2} - \frac 1 {2 a^4} \map \ln {a^2 - x^2} + \frac 1 {2 a^2 \paren {a^2 - x^2} } + C
      | c = Logarithm of Power and $x^2 > 0$
}}
{{eqn | r = \frac 1 {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by a squared minus x squared squared

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {a^2 - x^2}^2}
      | r = \int \paren {\frac 1 {a^4 x^2} + \frac 3 {4 a^5 \paren {a + x} } + \frac 3 {4 a^5 \paren {a - x} } + \frac 1 {4 a^4 \paren {a + x}^2} + \frac 1 {4 a^4 \paren {a - x}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} + \frac 3 {4 a^5} \int \frac {\d x} {a + x} + \frac 3 {4 a^5} \int \frac {\d x} {a - x} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \frac {-1} x + \frac 3 {4 a^5} \int \frac {\d x} {a + x} + \frac 3 {4 a^5} \int \frac {\d x} {a - x} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {a + x} + \frac 3 {4 a^5} \paren {-\ln \size {a - x} } + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {a + x} - \frac 3 {4 a^5} \ln \size {a - x} + \frac 1 {4 a^4} \frac {-1} {a + x} + \frac 1 {4 a^4} \frac 1 {a - x} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {\frac {a + x} {a - x} } + \frac {-1} {4 a^4 \paren {a + x} } + \frac 1 {4 a^4 \paren {a - x} } + C
      | c = Difference of Logarithms
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + \frac {-1} {4 a^4 \paren {a + x} } + \frac 1 {4 a^4 \paren {a - x} } + C
      | c = Sign of Quotient of Factors of Difference of Squares
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac x {2 a^4 \paren {a^2 - x^2} } + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by a squared minus x squared squared

:$\ds \int \frac {\d x} {x^3 \paren {a^2 - x^2}^2} = \frac {-1} {2 a^4 x^2} + \frac 1 {2 a^4 \paren {a^2 - x^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {a^2 - x^2}^2}
      | r = \int \paren {\frac 1 {a^4 x^3} + \frac 2 {a^6 x} + \frac {2 x} {a^6 \paren {a^2 - x^2} } + \frac x {a^4 \paren {a^2 - x^2}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {\d x} x + \frac 2 {a^6} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {a^2 - x^2}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \frac {-1} {2 x^2} + \frac 2 {a^6} \int \frac {\d x} x + \frac 2 {a^6} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x + \frac 2 {a^6} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x + \frac 2 {a^6} \paren {-\frac 1 2 \map \ln {a^2 - x^2} }  + \frac 1 {a^4} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
      | c = Primitive of $\dfrac x {a^2 - x^2}$
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 1 {a^6} \map \ln {a^2 - x^2}  + \frac 1 {a^4} \frac 1 {2 \paren {a^2 - x^2} }  + C
      | c = Primitive of $\dfrac x {\paren {a^2 - x^2}^2}$
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 1 {a^6} \map \ln {x^2} - \frac 1 {a^6} \map \ln {a^2 - x^2} + \frac 1 {2 a^4 \paren {a^2 - x^2} }  + C
      | c = Logarithm of Power and $x^2 > 0$
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 1 {2 a^4 \paren {a^2 - x^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {a^2 - x^2} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of a squared minus x squared

:$\ds \int \frac {\d x} {\paren {a^2 - x^2}^n} = \frac x {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {a^2 - x^2}^{n - 1} }$
for $x^2 > a^2$.

Proof:
Aiming for an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$
in order to use the technique of Integration by Parts, let:
{{begin-eqn}}
{{eqn | l = v
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = 1
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac 1 {\paren {a^2 - x^2}^{n - 1} }
      | c = 
}}
{{eqn | r = \paren {a^2 - x^2}^{-\paren {n - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = -\paren {n - 1} \paren {-2 x} \paren {a^2 - x^2}^{-\paren {n - 1} - 1}
      | c = Power Rule for Derivatives and Chain Rule for Derivatives
}}
{{eqn | r = 2 \paren {n - 1} x \paren {a^2 - x^2}^{-n}
      | c = 
}}
{{eqn | r = \frac {\paren {2 n - 2} x} {\paren {a^2 - x^2}^n}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a^2 - x^2}^{n - 1} }
      | r = \frac 1 {\paren {a^2 - x^2}^{n - 1} } x - \int x \frac {\paren {2 n - 2} x} {\paren {a^2 - x^2}^n} \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac x {\paren {a^2 - x^2}^{n - 1} } - \paren {2 n - 2} \int \frac {x^2} {\paren {a^2 - x^2}^n} \rd x
      | c = simplifying
}}
{{eqn | r = \frac x {\paren {a^2 - x^2}^{n - 1} } - \paren {2 n - 2} \int \frac {x^2 - a^2 + a^2} {\paren {a^2 - x^2}^n} \rd x
      | c = 
}}
{{eqn | r = \frac x {\paren {a^2 - x^2}^{n - 1} } - \paren {2 n - 2} \int \frac {-\paren {a^2 - x^2} } {\paren {a^2 - x^2}^n} \rd x - \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {a^2 - x^2}^n}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac x {\paren {a^2 - x^2}^{n - 1} } + \paren {2 n - 2} \int \frac {\d x} {\paren {a^2 - x^2}^{n - 1} } - \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {a^2 - x^2}^n} 
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {1 - \paren {2 n - 2} } \int \frac {\d x} {\paren {a^2 - x^2}^{n - 1} }
      | r = \frac x {\paren {a^2 - x^2}^{n - 1} } + \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {a^2 - x^2}^n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\paren {a^2 - x^2}^n}
      | r = \frac x {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {a^2 - x^2}^{n - 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Power of a squared minus x squared

:$\ds \int \frac {x \rd x} {\paren {a^2 - x^2}^n} = \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} }$
for $x^2 < a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a^2 - x^2
      | c = 
}}
{{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 {a^2 - x^2}^n}
      | r = \int \frac {\d z} {- 2 z^n}
      | c = Integration by Substitution
}}
{{eqn | r = \frac {-1} 2 \int z^{-n} \rd z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {-1} 2 \frac {z^{-n + 1} } {-n + 1} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac 1 {2 \paren {n - 1} z^{n - 1} } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} } + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Power of a squared minus x squared

:$\ds \int \frac {\d x} {x \paren {a^2 - x^2}^n} = \frac 1 {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {a^2 - x^2}^{n - 1} }$
for $x^2 < a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a^2 - x^2
      | c = 
}}
{{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 {a^2 - x^2}^n}
      | r = \int \frac {\d z} {-2 x^2 z^n}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\d z} {\paren {z - a^2} z^n}
      | c = Primitive of Constant Multiple of Function and substituting for $x^2$
}}
{{end-eqn}}
From Primitive of $x^m \paren {a x + b}^n$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Let:
{{begin-eqn}}
{{eqn | l = a
      | o = :=
      | r = 1
      | c = 
}}
{{eqn | l = b
      | o = :=
      | r = -a^2
      | c = 
}}
{{eqn | l = x
      | o = :=
      | r = z
      | c = 
}}
{{eqn | l = m
      | o = :=
      | r = -n
      | c = 
}}
{{eqn | l = n
      | o = :=
      | r = -1
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2}^n}
      | r = \frac 1 2 \int \frac {\d z} {\paren {z - a^2} z^n}
      | c = 
}}
{{eqn | r = \frac 1 2 \paren {\frac {z^{-n + 1} \paren {z - a^2}^{-1 + 1} } {\paren {-n + 1} \paren {-a^2} } - \frac {\paren {-n + \paren {-1} + 2} } {\paren {-n + 1} \paren {-a^2} } \int z^{-n + 1} \paren {z - a^2}^{-1} \rd z}
      | c = 
}}
{{eqn | r = \frac 1 {2 \paren {n - 1} a^2 z^{n-1} } + \frac 1 {2 a^2} \int \frac {\d z} {z^{n - 1} \paren {z - a^2} }
      | c = simplification
}}
{{eqn | r = \frac 1 {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \paren {a^2 - x^2}^{n - 1} }
      | c = substituting for $z$ and $\d z$
}}
{{eqn | r = \frac 1 {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {a^2 - x^2}^{n - 1} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x over Power of a squared minus x squared

:$\ds \int \frac {x^m \rd x} {\paren {a^2 - x^2}^n} = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^{n - 1} }$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^m \rd x} {\paren {a^2 - x^2}^n}
      | r = \int \frac {x^{m - 2} \paren {x^2} \rd x} {\paren {a^2 - x^2}^n}
      | c = 
}}
{{eqn | r = \int \frac {x^{m - 2} \paren {a^2 + x^2 - a^2} \rd x} {\paren {a^2 - x^2}^n}
      | c = 
}}
{{eqn | r = \int \frac {x^{m - 2} \paren {a^2 - \paren {a^2 - x^2} } \rd x} {\paren {a^2 - x^2}^n}
      | c = 
}}
{{eqn | r = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \paren {a^2 - x^2} \rd x} {\paren {a^2 - x^2}^n}
      | c = Linear Combination of Primitives
}}
{{eqn | r = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^{n - 1} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of x by Power of a squared minus x squared

:$\ds \int \frac {\d x} {x^m \paren {a^2 - x^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {a^2 - x^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {a^2 - x^2}^n}$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^m \paren {a^2 - x^2}^{n - 1} }
      | r = \int \frac {\paren {a^2 - x^2} \rd x} {x^m \paren {a^2 - x^2}^{n - 1} \paren {a^2 - x^2} }
      | c = 
}}
{{eqn | r = \int \frac {\paren {a^2 - x^2} \rd x} {x^m \paren {a^2 - x^2}^{\paren {n - 1} + 1} }
      | c = 
}}
{{eqn | r = a^2 \int \frac {\d x} {x^m \paren {a^2 - x^2}^n} - \int \frac {x^2 \rd x} {x^m \paren {a^2 - x^2}^n}
      | c = Linear Combination of Primitives
}}
{{eqn | r = a^2 \int \frac {\d x} {x^m \paren {a^2 - x^2}^n} - \int \frac {\d x} {x^{m - 2} \paren {a^2 - x^2}^n}
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = a^2 \int \frac {\d x} {x^m \paren {a^2 - x^2}^n} 
      | r = \int \frac {\d x} {x^m \paren {a^2 - x^2}^{n - 1} } + \int \frac {\d x} {x^{m - 2} \paren {a^2 - x^2}^n}
      | c = changing sides
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x^m \paren {a^2 - x^2}^n} 
      | r = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {a^2 - x^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {a^2 - x^2}^n}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Root of x squared plus 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 plus 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 plus a squared

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

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^2 \sqrt {x^2 + a^2} }
      | r = \int \frac {\d z} {2 z \sqrt z \sqrt {z + a^2} }
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\d z} {z^{3/2} \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 = \frac 3 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^{3/2} \sqrt {z + a^2} }
      | r = \frac {-\sqrt{z + a^2} } {2 \paren {\paren {\frac 3 2} - 1} a^2 z^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} a^2} \int \frac {\d z} {z^{\paren {\frac 3 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} } {a^2 z^{1/2} } - 0 + C
      | c = simplifying
}}
{{eqn | r = -\frac {\sqrt {x^2 + a^2} } {a^2 x} + C
      | c = substituting back for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by Root of x squared plus a squared

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

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^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} a^2 z^{2 - 1} } - \frac {2 \paren 2 - 3} {2 \paren {2 \paren 2 - 2} 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} \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 x by Root of x squared plus 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}}