avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Reciprocal of x cubed plus a cubed squared

:$\ds \int \frac {\d x} {\paren {x^3 + a^3}^2} = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 1 {9 a^5} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

Proof:
{{begin-eqn}}
{{eqn | l = z
      | r = x^3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 3 x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\paren {x^3 + a^3}^2}
      | r = \int \frac {\d z} {3 x^2 \paren {z + a^3}^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 3 \int \frac {\d z} {z^{2/3} \paren {z + a^3}^2}
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
Then from Primitive of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$:
:$\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$
Here we have $a = 1, b = 0, m = \dfrac 2 3, p = 1, q = a^3, n = 2$.
So:
{{begin-eqn}}
{{eqn | o = 
      | r = \frac 1 3 \int \frac {\d z} {z^{2/3} \paren {z + a^3}^2}
      | c = 
}}
{{eqn | r = \frac 1 3 \paren {\frac {-1} {\paren 1 \paren {-a^3} } \paren {\frac 1 {z^{-1/3} \paren {z + a^3}^1} + \paren {\frac 2 3 + 2 - 2} \int \frac {\d z} {\paren z^{2/3} \paren {z + a^3}^1} } }
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$
}}
{{eqn | r = \frac 1 {3 a^3 z^{-1/3} \paren {z + a^3} } + \frac 2 {9 a^3} \int \frac {\d z} {z^{2/3} \paren {z + a^3} }
      | c = simplifying
}}
{{eqn | r = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 2 {9 a^3} \int \frac {3 x^2 \rd x} {x^2 \paren {x^3 + a^3} }
      | c = substituting for $z$
}}
{{eqn | r = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 2 {3 a^3} \int \frac {\d x} {\paren {x^3 + a^3} }
      | c = simplifying
}}
{{eqn | r = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 2 {3 a^3} \paren {\frac 1 {6 a^2} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }
      | c = Primitive of $\dfrac 1 {\paren {x^3 + a^3} }$
}}
{{eqn | r = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 1 {9 a^5} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x cubed plus a cubed squared

:$\ds \int \frac {x \rd x} {\paren {x^3 + a^3}^2} = \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {18 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {3 a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

Proof:
First a lemma:
Theorem: Primitive of x squared over x cubed plus a cubed squared

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

Proof:
{{begin-eqn}}
{{eqn | l = z
      | r = x^3 + a^3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 3 x^2
      | c = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x^2 \rd x} {\paren {x^3 + a^3}^2}
      | r = \int \frac {\d z} {3 z^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac {-1} {3 z} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {3 \paren {x^3 + a^3} } + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by x cubed plus a cubed squared

:$\ds \int \frac {\d x} {x \paren {x^3 + a^3}^2} = \frac 1 {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^6} \ln \size {\frac {x^3} {x^3 + a^3} }$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^3 + a^3}^2}
      | r = \int \frac {a^3 \rd x} {a^3 x \paren {x^3 + a^3}^2}
      | c = multiplying top and bottom by $a^3$
}}
{{eqn | r = \int \frac {\paren {x^3 + a^3 - x^3} \rd x} {a^3 x \paren {x^3 + a^3}^2}
      | c = 
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\paren {x^3 + a^3} \rd x} {x \paren {x^3 + a^3}^2} - \frac 1 {a^3} \int \frac {x^3 \rd x} {x \paren {x^3 + a^3}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\d x} {x \paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x^2 \rd x} {\paren {x^3 + a^3}^2}
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^3} \paren {\frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } } - \frac 1 {a^3} \int \frac {x^2 \rd x} {\paren {x^3 + a^3}^2}
      | c = Primitive of $\dfrac 1 {x \paren {x^3 + a^3} }$
}}
{{eqn | r = \frac 1 {a^3} \paren {\frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } } - \frac 1 {a^3} \paren {\frac {-1} {3 \paren {x^3 + a^3} } }
      | c = Primitive of $\dfrac {x^2} {\paren {x^3 + a^3}^2}$
}}
{{eqn | r = \frac 1 {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^6} \ln \size {\frac {x^3} {x^3 + a^3} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by x cubed plus a cubed squared

:$\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3}^2} = \frac {-1} {a^6 x} - \frac {x^2} {3 a^6 \paren {x^3 + a^3} } - \frac 4 {3 a^6} \int \frac {x \rd x} {x^3 + a^3}$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {x^3 + a^3}^2}
      | r = \int \frac {a^3 \rd x} {a^3 x^2 \paren {x^3 + a^3}^2}
      | c = multiplying top and bottom by $a^3$
}}
{{eqn | r = \int \frac {\paren {x^3 + a^3 - x^3} \rd x} {a^3 x^2 \paren {x^3 + a^3}^2}
      | c = 
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\paren {x^3 + a^3} \rd x} {x^2 \paren {x^3 + a^3}^2} - \frac 1 {a^3} \int \frac {x^3 \rd x} {x^2 \paren {x^3 + a^3}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\d x} {x^2 \paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3}^2}
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^3} \paren {\frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} } } - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3}^2}
      | c = Primitive of $\dfrac 1 {x^2 \paren {x^3 + a^3} }$: Lemma
}}
{{eqn | r = \frac {-1} {a^6 x} - \frac 1 {a^6} \int \frac {x \rd x} {\paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3}^2}
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^6 x} - \frac 1 {a^6} \int \frac {x \rd x} {\paren {x^3 + a^3} } - \frac 1 {a^3} \paren {\frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} } }
      | c = Primitive of $\dfrac x {\paren {x^3 + a^3}^2}$: Lemma
}}
{{eqn | r = \frac {-1} {a^6 x} - \frac {x^2} {3 a^6 \paren {x^3 + a^3} } - \frac 4 {3 a^6} \int \frac {x \rd x} {x^3 + a^3}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x over x cubed plus a cubed

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^m \rd x} {x^3 + a^3}
      | r = \int \frac {x^{m - 3} \paren {x^3} \rd x} {x^3 + a^3}
      | c = 
}}
{{eqn | r = \int \frac {x^{m - 3} \paren {x^3 + a^3 - a^3} \rd x} {x^3 + a^3}
      | c = 
}}
{{eqn | r = \int \frac {x^{m - 3} \paren {x^3 + a^3} \rd x} {x^3 + a^3} - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int x^{m - 3} \rd x - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}
      | c = simplification
}}
{{eqn | r = \frac {x^{m - 2} } {m - 2} - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}
      | c = Primitive of Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of x by x cubed plus a cubed

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^n \paren {x^3 + a^3} }
      | r = \int \frac {a^3 \rd x} {a^3 x^n \paren {x^3 + a^3} }
      | c = 
}}
{{eqn | r = \int \frac {\paren {x^3 + a^3 - x^3} \rd x} {a^3 x^n \paren {x^3 + a^3} }
      | c = 
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\paren {x^3 + a^3} \rd x} {x^n \paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x^3 \rd x} {x^n \paren {x^3 + a^3} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\d x} {x^n} - \frac 1 {a^3} \int \frac {x^3 \rd x} {x^n \paren {x^3 + a^3} }
      | c = simplification
}}
{{eqn | r = \frac {-1} {a^3 \paren {n - 1} x^{n - 1} } - \frac 1 {a^3} \int \frac {\d x} {x^{n - 3} \paren {x^3 + a^3} }
      | c = Primitive of Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x fourth plus a fourth

:$\ds \int \frac {x \rd x} {x^4 + a^4} = \frac 1 {2 a^2} \arctan \frac {x^2} {a^2}$

Proof:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {x^4 + a^4}
      | r = \int \frac {\d z} {2 \paren {z^2 + \paren {a^2}^2} }
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {2 a^2} \arctan \frac z {a^2} + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \frac 1 {2 a^2} \arctan \frac {x^2} {a^2} + C
      | c = substituting back for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over x fourth plus a fourth

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

Proof:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} x^4
      | r = 4 x^3
      | c = Primitive of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x^3 \rd x} {x^4 + a^4}
      | r = \frac 1 4 \ln \size {x^4 + a^4} + C
      | c = Primitive of Function under its Derivative
}}
{{eqn | r = \frac {\map \ln {x^4 + a^4} } 4 + C
      | c = Absolute Value of Even Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over x fourth plus a fourth

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

Proof:
From Primitive of $\dfrac {x^{n - 1} } {x^n + a^n}$:
:$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$
So:
{{begin-eqn}}
{{eqn | l = \int \frac {x^3 \rd x} {x^4 + a^4}
      | r = \frac 1 4 \ln \size {x^4 + a^4} + C
      | c = Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 4$
}}
{{eqn | r = \frac {\map \ln {x^4 + a^4} } 4 + C
      | c = Absolute Value of Even Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by x fourth plus a fourth

:$\ds \int \frac {\d x} {x \paren {x^4 + a^4} } = \frac 1 {4 a^4} \map \ln {\frac {x^4} {x^4 + a^4} }$

Proof:
From Primitive of Reciprocal of x by Power of x plus Power of a:
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
So:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^4 + a^4} }
      | r = \frac 1 {4 a^4} \ln \size {\frac {x^4} {x^4 + a^4} } + C
      | c = Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$ with $n = 4$
}}
{{eqn | r = \frac 1 {4 a^4} \map \ln {\frac {x^4} {x^4 + a^4} } + C
      | c = Absolute Value of Even Power
}}
{{end-eqn}}
directly.
{{qed}}
Theorem: Primitive of Reciprocal of x squared by x fourth plus a fourth

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

Proof:
{{begin-eqn}}
{{eqn | r = \int \frac {\d x} {x^2 \paren {x^4 + a^4} }
      | o = 
      | c = 
}}
{{eqn | r = \int \frac {a^4 \rd x} {a^4 x^2 \paren {x^4 + a^4} }
      | c = multiplying top and bottom by $a^4$
}}
{{eqn | r = \int \frac {\paren {x^4 + a^4 - x^4} \rd x} {a^4 x^2 \paren {x^4 + a^4} }
      | c = 
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\paren {x^4 + a^4} \rd x} {x^2 \paren {x^4 + a^4} } - \frac 1 {a^4} \int \frac {x^4 \rd x} {x^2 \paren {x^4 + a^4} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} - \frac 1 {a^4} \int \frac {x^2 \rd x} {x^4 + a^4}
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^4} \paren {\frac {-1} x} - \frac 1 {a^4} \int \frac {x^2 \rd x} {x^4 + a^4}
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {a^4 x} - \frac 1 {a^4} \paren {\frac 1 {4 a \sqrt 2} \map \ln {\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} } - \frac 1 {2 a \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} } }
      | c = Primitive of $\dfrac {x^2} {x^4 + a^4}$
}}
{{eqn | r = \frac {-1} {a^4 x} - \frac {-1} {4 a^5 \sqrt 2} \map \ln {\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} } + \frac 1 {2 a^5 \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by x fourth plus a fourth

:$\ds \int \frac {\d x} {x^3 \paren {x^4 + a^4} } = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^6} \arctan \frac {x^2} {a^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {x^4 + a^4} }
      | r = \int \frac {a^4 \rd x} {a^4 x^3 \paren {x^4 + a^4} }
      | c = multiplying top and bottom by $a^4$
}}
{{eqn | r = \int \frac {\paren {x^4 + a^4 - x^4} \rd x} {a^4 x^3 \paren {x^4 + a^4} }
      | c = 
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\paren {x^4 + a^4} \rd x} {x^3 \paren {x^4 + a^4} } - \frac 1 {a^4} \int \frac {x^4 \rd x} {x^3 \paren {x^4 + a^4} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^3} - \frac 1 {a^4} \int \frac {x \rd x} {x^4 + a^4}
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^4} \paren {\frac {-1} {2 x} } - \frac 1 {a^4} \int \frac {x \rd x} {x^4 + a^4}
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} - \frac 1 {a^4} \paren {\frac 1 {2 a^2} \arctan \frac {x^2} {a^2} }
      | c = Primitive of $\dfrac x {x^4 + a^4}$
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^6} \arctan \frac {x^2} {a^2}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x fourth minus a fourth

:$\ds \int \frac {\d x} {x^4 - a^4} = \frac 1 {4 a^3} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a^3} \arctan \frac x a$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^4 - a^4}
      | r = \int \frac {\d x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Difference of Two Squares
}}
{{eqn | r = \int \frac {a^2 \rd x} {a^2 \paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = multiplying top and bottom by $a^2$
}}
{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 \paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = 
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \d x} {\paren {x^2 + a^2} \paren {x^2 - a^2} } - \frac 1 {a^2} \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\paren {x^2 - a^2 + a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Difference of Two Squares
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\paren {x^2 - a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Linear Combination of Primitives
}}
{{eqn | o = 
      | ro= -
      | r = \frac 1 {a^2} \int \frac {a^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = 
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} {x^2 + a^2} - \int \frac {\d x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = simplification
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} {x^2 + a^2} - \int \frac {\d x} {x^4 - a^4}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = 2 \int \frac {\d x} {x^4 - a^4}
      | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} {x^2 + a^2} 
      | c = gathering terms
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x^4 - a^4}
      | r = \frac 1 {2 a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {2 a^2} \int \frac {\d x} {x^2 + a^2} 
      | c = gathering terms
}}
{{eqn | r = \frac 1 {4 a^3} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a^2} \int \frac {\d x} {x^2 + a^2} 
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {4 a^3} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a^3} \arctan \frac x a
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x fourth minus a fourth

:$\ds \int \frac {x \rd x} {x^4 - a^4} = \frac 1 {4 a^2} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } + C$

Proof:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {x^4 - a^4}
      | r = \int \frac {\d z} {2 \paren {z^2 - \paren {a^2}^2} }
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 {2 a^2} \ln \size {\frac {z - a^2} {z + a^2} } } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {4 a^2} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } + C
      | c = substituting back for $z$ and simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x fourth minus a fourth

:$\ds \int \frac {x^2 \rd x} {x^4 - a^4} = \frac 1 {4 a} \ln \size {\frac {x - a} {x + a} } + \frac 1 {2 a} \arctan \frac x a + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {x^4 - a^4}
      | r = \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Difference of Two Squares
}}
{{eqn | r = \int \frac {\paren {x^2 + a^2 - a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = 
}}
{{eqn | r = \int \frac {\paren {x^2 + a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} } + \int \frac {-a^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {-a^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = simplifying
}}
{{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\paren {x^2 - a^2 - x^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Difference of Two Squares
}}
{{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\paren {x^2 - a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = Linear Combination of Primitives
}}
{{eqn | o = 
      | ro= -
      | r = \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = 
}}
{{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\d x} {x^2 + a^2} - \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }
      | c = simplification
}}
{{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\d x} {x^2 + a^2} - \int \frac {x^2 \rd x} {x^4 - a^4}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = 2 \int \frac {x^2 \rd x} {x^4 - a^4}
      | r = \int \frac {\d x} {x^2 - a^2} - \int \frac {\d x} {x^2 + a^2} 
      | c = gathering terms
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x^4 - a^4}
      | r = \frac 1 2 \int \frac {\d x} {x^2 - a^2} - \frac 1 2 \int \frac {\d x} {x^2 + a^2} 
      | c = gathering terms
}}
{{eqn | r = \frac 1 {4 a} \ln \size {\frac {x - a} {x + a} } - \frac 1 2 \int \frac {\d x} {x^2 + a^2} 
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {4 a} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a} \arctan \frac x a
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over x fourth minus a fourth

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

Proof:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} x^4
      | r = 4 x^3
      | c = Primitive of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x^3 \rd x} {x^4 - a^4}
      | r = \frac {\ln \size {x^4 - a^4} } 4 + C
      | c = Primitive of Function under its Derivative
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by x fourth minus a fourth

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

Proof:
From Primitive of $\dfrac 1 {x \paren {x^n - a^n} }$:
:$\ds \int \frac {\d x} {x \paren {x^n - a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n - a^n} {x^n} } + C$
Setting $n = 4$:
:$\ds \int \frac {\d x} {x \paren {x^4 - a^4} } = \frac 1 {4 a^4} \ln \size {\frac {x^4 - a^4} {x^4} } + C$
directly.
{{qed}}
Theorem: Primitive of Reciprocal of x squared by x fourth minus a fourth

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {x^4 - a^4} }
      | r = \int \frac {a^4 \rd x} {a^4 x^2 \paren {x^4 - a^4} }
      | c = multiplying top and bottom by $a^4$
}}
{{eqn | r = \int \frac {\paren {x^4 + a^4 - x^4} \rd x} {a^4 x^2 \paren {x^4 - a^4} }
      | c = 
}}
{{eqn | r = \int \frac {\paren {x^4 - \paren {x^4 - a^4} } \rd x} {a^4 x^2 \paren {x^4 - a^4} }
      | c = 
}}
{{eqn | r = \frac {-1} {a^4} \int \frac {\paren {x^4 - a^4} \rd x} {x^2 \paren {x^4 + a^4} } + \frac 1 {a^4} \int \frac {x^4 \rd x} {x^2 \paren {x^4 - a^4} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {-1} {a^4} \int \frac {\d x} {x^2} + \frac 1 {a^4} \int \frac {x^2 \rd x} {x^4 - a^4}
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^4} \paren {\frac {-1} x} + \frac 1 {a^4} \int \frac {x^2 \rd x} {x^4 - a^4}
      | c = Primitive of Power
}}
{{eqn | r = \frac 1 {a^4 x} + \frac 1 {a^4} \paren {\frac 1 {4 a} \ln \size {\frac {x - a} {x + a} } + \frac 1 {2 a} \arctan \frac x a} + C
      | c = Primitive of $\dfrac {x^2} {x^4 - a^4}$
}}
{{eqn | r = \frac 1 {a^4 x} + \frac 1 {4 a^5} \ln \size {\frac {x - a} {x + a} } + \frac 1 {2 a^5} \arctan \frac x a + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by x fourth minus a fourth

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {x^4 - a^4} }
      | r = \int \frac {a^4 \rd x} {a^4 x^3 \paren {x^4 - a^4} }
      | c = multiplying top and bottom by $a^4$
}}
{{eqn | r = \int \frac {\paren {x^4 + a^4 - x^4} \rd x} {a^4 x^3 \paren {x^4 - a^4} }
      | c = 
}}
{{eqn | r = \int \frac {\paren {x^4 - \paren {x^4 - a^4} } \rd x} {a^4 x^3 \paren {x^4 - a^4} }
      | c = 
}}
{{eqn | r = \frac {-1} {a^4} \int \frac {\paren {x^4 - a^4} \rd x} {x^3 \paren {x^4 - a^4} } + \frac 1 {a^4} \int \frac {x^4 \rd x} {x^3 \paren {x^4 - a^4} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {-1} {a^4} \int \frac {\d x} {x^3} + \frac 1 {a^4} \int \frac {x \rd x} {x^4 - a^4}
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^4} \paren {\frac {-1} {2 x} } + \frac 1 {a^4} \int \frac {x \rd x} {x^4 - a^4}
      | c = Primitive of Power
}}
{{eqn | r = \frac 1 {2 a^4 x^2} + \frac 1 {a^4} \paren {\frac 1 {4 a^2} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } } + C
      | c = Primitive of $\dfrac x {x^4 - a^4}$
}}
{{eqn | r = \frac 1 {2 a^4 x^2} + \frac 1 {4 a^6} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}