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

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

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

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

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

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

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

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

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

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

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

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2} }
      | r = \int \frac {a^2 \rd x} {a^2 x \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 x \paren {x^2 + a^2} }
      | c = adding and subtracting $x^2$
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \rd x} {x \paren {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x^2 \rd x} {x \paren {x^2 + a^2} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2}
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \paren {\frac 1 2 \map \ln {x^2 + a^2} } + C
      | c = Primitive of $\dfrac x {x^2 + a^2}$
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by x squared plus a squared

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

Proof:
From Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$:
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
So:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2} }
      | r = \frac 1 {2 a^2} \ln \size {\frac {x^2} {x^2 + a^2} } + C
      | c = Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$ with $n = 2$
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C
      | c = Absolute Value of Even Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by x squared plus a squared

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

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

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

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

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = a \tan \theta
      | c = 
}}
{{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 {x^2 + a^2}^2}
      | r = \int \frac {a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac {a \sec^2 \theta} {\paren {a^2 \sec^2 \theta}^2} \rd \theta
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\d \theta} {\sec^2 \theta}
      | c = simplification
}}
{{eqn | r = \frac 1 {a^3} \int \cos^2 \theta \rd \theta
      | c = {{Defof|Real Secant Function}}
}}
{{eqn | r = \frac 1 {a^3} \paren {\frac \theta 2 + \frac {\sin 2 \theta} 4} + C
      | c = Primitive of Square of Cosine Function
}}
{{eqn | r = \frac \theta {2 a^3} + \frac 1 {2 a^3} \frac {\tan \theta} {1 + \tan^2 \theta} + C
      | c = Tangent Half-Angle Substitution for Sine
}}
{{eqn | r = \frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a + C
      | c = substituting for $\theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x squared plus a squared squared

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

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

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = a \tan \theta
      | c = 
}}
{{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 {x^2 \rd x} {\paren {x^2 + a^2}^2}
      | r = \int \frac {a^2 \tan^2 \theta a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac {a^3 \tan^2 \theta \sec^2 \theta} {a^4 \sec^4 \theta} \rd \theta
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \frac 1 a \int \frac {\tan^2 \theta} {\sec^2 \theta} \rd \theta
      | c = simplification
}}
{{eqn | r = \frac 1 a \int \tan^2 \theta \cos^2 \theta \rd \theta
      | c = {{Defof|Real Secant Function}}
}}
{{eqn | r = \frac 1 a \int \frac {\sin^2 \theta} {\cos^2 \theta} \cos^2 \theta \rd \theta
      | c = {{Defof|Real Tangent Function}}
}}
{{eqn | r = \frac 1 a \int \sin^2 \theta \rd \theta
      | c = simplification
}}
{{eqn | r = \frac \theta {2 a} - \frac {\sin 2 \theta} {4 a} + C
      | c = Primitive of Square of Sine Function
}}
{{eqn | r = \frac \theta {2 a}  - \frac 1 {2 a} \frac {\tan \theta} {1 + \tan^2 \theta} + C
      | c = Tangent Half-Angle Substitution for Sine
}}
{{eqn | r = -\frac x {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C
      | c = substituting for $\theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x squared plus a squared squared

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

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

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

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

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

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

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2}^2}
      | r = \int \paren {\frac 1 {a^4 x} - \frac x {a^4 \paren {x^2 + a^2} } - \frac x {a^2 \paren {x^2 + a^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} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C
      | c = Primitive of $\dfrac x {x^2 + a^2}$
}}
{{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \paren {-\frac 1 {2 \paren {x^2 + a^2} } } + C
      | c = Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$
}}
{{eqn | r = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by x squared plus a squared squared

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2}
      | r = \int \paren {\frac 1 {a^4 x^2} - \frac 1 {a^4 \paren {x^2 + a^2} } - \frac 1 {a^2 \paren {x^2 + a^2}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \frac {-1} x - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2} + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac 1 {a^4 x} - \frac 1 {a^4} \paren {\frac 1 a \arctan \frac x a} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2} + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = -\frac 1 {a^4 x} - \frac 1 {a^5} \arctan \frac x a - \frac 1 {a^2} \paren {\frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a} + C
      | c = Primitive of $\dfrac 1 {\paren {x^2 + a^2}^2}$
}}
{{eqn | r = -\frac 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by x squared plus a squared squared

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

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

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

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 {x^2 + a^2}^{n - 1} }
      | c = 
}}
{{eqn | r = \paren {x^2 + a^2}^{-\paren {n - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = -2 \paren {n - 1} x \paren {x^2 + a^2}^{-\paren {n - 1} - 1}
      | c = Power Rule for Derivatives and Chain Rule for Derivatives
}}
{{eqn | r = -2 \paren {n - 1} x \paren {x^2 + a^2}^{-n}
      | c = 
}}
{{eqn | r = \frac {-\paren {2 n - 2} x} {\paren {x^2 + a^2}^n}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {x^2 + a^2}^{n - 1} }
      | r = \frac 1 {\paren {x^2 + a^2}^{n - 1} } x - \int x \frac {-\paren {2 n - 2} x} {\paren {x^2 + a^2}^n} \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac x {\paren {x^2 + a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {x^2} {\paren {x^2 + a^2}^n} \rd x
      | c = simplifying
}}
{{eqn | r = \frac x {\paren {x^2 + a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {x^2 + a^2 - a^2} {\paren {x^2 + a^2}^n} \rd x
      | c = 
}}
{{eqn | r = \frac x {\paren {x^2 + a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {x^2 + a^2} {\paren {x^2 + a^2}^n} \rd x - \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {x^2 + a^2}^n}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac x {\paren {x^2 + a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {\d x} {\paren {x^2 + a^2}^{n - 1} } - \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {x^2 + a^2}^n} 
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {1 - \paren {2 n - 2} } \int \frac {\d x} {\paren {x^2 + a^2}^{n - 1} }
      | r = \frac x {\paren {x^2 + a^2}^{n - 1} } - \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {x^2 + a^2}^n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\paren {x^2 + a^2}^n}
      | r = \frac x {2 \paren {n - 1} a^2 \paren {x^2 + a^2}^{n - 1} } + \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {x^2 + a^2}^{n - 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Power of x squared plus a squared

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

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2 + a^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 {x^2 + a^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 {x^2 + a^2}^{n - 1} } + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}