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

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\sqrt {a x^2 + b x + c} } x \rd x = \sqrt {a x^2 + b x + c} + \frac b 2 \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$

Proof:
{{begin-eqn}}
{{eqn | r = \int \frac {\sqrt {a x^2 + b x + c} } x \rd x
      | o = 
}}
{{eqn | r = \int \frac {a x^2 + b x + c} {x \sqrt {a x^2 + b x + c} } \rd x
      | c = multiplying top and bottom by $\sqrt {a x^2 + b x + c}$
}}
{{eqn | r = a \int \frac {x^2 \rd x} {x \sqrt {a x^2 + b x + c} } + b \int \frac {x \rd x} {x \sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = a \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} } + b \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{eqn | r = a \paren {\frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\dfrac x {\sqrt {a x^2 + b x + c} }$
}}
{{eqn | o = 
      | ro= +
      | r = b \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \sqrt {a x^2 + b x + c} + \frac b 2 \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a x squared plus b x plus c over x squared

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\sqrt {a x^2 + b x + c} } {x^2} \rd x = \frac {-\sqrt {a x^2 + b x + c} } x + a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + \frac b 2 \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$

Proof:
{{begin-eqn}}
{{eqn | r = \int \frac {\sqrt {a x^2 + b x + c} } {x^2} \rd x
      | o = 
}}
{{eqn | r = \int \frac {a x^2 + b x + c} {x^2 \sqrt {a x^2 + b x + c} } \rd x
      | c = multiplying top and bottom by $\sqrt {a x^2 + b x + c}$
}}
{{eqn | r = a \int \frac {x^2 \rd x} {x^2 \sqrt {a x^2 + b x + c} } + b \int \frac {x \rd x} {x^2 \sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x^2 \sqrt {a x^2 + b x + c} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + b \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x^2 \sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{eqn | r = c \paren {-\frac {\sqrt {a x^2 + b x + c} } {c x} - \frac b {2 c} \frac {\d x} {x \sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {a x^2 + b x + c} }$
}}
{{eqn | o = 
      | ro= +
      | r = a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + b \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac {-\sqrt {a x^2 + b x + c} } x + a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + \frac b 2 \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Cube of Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + C$

Proof:
For $a > 0$:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | r = \int \frac {\d x} {\paren {\sqrt {\frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a} } }^3}
      | c = Completing the Square
}}
{{eqn | r = \int \frac {8 a \sqrt a \rd x} {\paren {\sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} }^3}
      | c = simplifying
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = 2 a x + b
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a
      | c = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {8 a \sqrt a \rd x} {\paren {\sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} }^3}
      | r = \int \frac {8 a \sqrt a \rd z} {2 a \paren {\sqrt {z^2 + 4 a c - b^2} }^3}
      | c = Integration by Substitution
}}
{{eqn | r = 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Let $4 a c - b^2 > 0$.
Then:
{{begin-eqn}}
{{eqn | l = 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}
      | r = 4 \sqrt a \paren {\frac z {\paren {4 a c - b^2} \sqrt {z^2 + 4 a c - b^2} } } + C
      | c = Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$
}}
{{end-eqn}}
Let $4 a c - b^2 < 0$.
Then:
{{begin-eqn}}
{{eqn | l = 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}
      | r = 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 - \paren {b^2 - 4 a c} } }^3}
      | c = 
}}
{{eqn | r = 4 \sqrt a \paren {\frac {-z} {\paren {b^2 - 4 a c} \sqrt {z^2 - \paren {b^2 - 4 a c} } } } + C
      | c = Primitive of $\dfrac 1 {\paren {\sqrt {x^2 - a^2} }^3}$
}}
{{eqn | r = 4 \sqrt a \paren {\frac z {\paren {4 a c - b^2} \sqrt {z^2 + 4 a c - b^2} } } + C
      | c = simplifying
}}
{{end-eqn}}
Thus in both cases the same result applies, and so:
{{begin-eqn}}
{{eqn | l = 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}
      | r = \frac {4 \sqrt a \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} } + C
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {4 \sqrt a \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {4 a \paren {a x^2 + b x + c} } } + C
      | c = Completing the Square
}}
{{eqn | r = \frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
{{finish|Similarly for $a < 0$, unless there's a quick way to cover both cases}}
{{qed}}
Theorem: Primitive of x over Cube of Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } + C$

Proof:
First:
{{begin-eqn}}
{{eqn | l = z
      | r = a x^2 + b x + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a x + b
      | c = Derivative of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | r = \frac 1 {2 a} \int \frac {2 a x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b {2 a} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} 
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\d z} {z^{3/2} } - \frac b {2 a} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} 
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {2 a} \paren {\frac {-2} {\sqrt z} } - \frac b {2 a} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = Primitive of Power
}}
{{eqn | r = \frac 1 {a \sqrt z} - \frac b {2 a} \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } } + C
      | c = Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$
}}
{{eqn | r = \frac 1 {a \sqrt {a x^2 + b x + c} } - \frac b {2 a} \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } } + C
      | c = substituting for $z$
}}
{{eqn | r = \frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over Cube of Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x^2 \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {\paren {2 b^2 - 4 a c} x + 2 b c} {a \paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$

Proof:
{{begin-eqn}}
{{eqn | r = \int \frac {x^2 \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | o = 
      | c = 
}}
{{eqn | r = \int \frac {a x^2 \rd x} {a \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = multiplying top and bottom by $a$
}}
{{eqn | r = \int \frac {\paren {a x^2 + b x + c - b x - c} \rd x} {a \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {\paren {a x^2 + b x + c} \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b a \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac c a \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac b a \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac c a \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = simplifying
}}
{{eqn | r = \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac b a \paren {\frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\dfrac x {\paren {\sqrt {a x^2 + b x + c} }^3}$
}}
{{eqn | o = 
      | ro= -
      | r = \frac c a \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$
}}
{{eqn | r = \frac {\paren {2 b^2 - 4 a c} x + 2 b c} {a \paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Cube of Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3} = \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}$

Proof:
{{begin-eqn}}
{{eqn | r = \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3}
      | o = 
      | c = 
}}
{{eqn | r = \int \frac {c \rd x} {c x \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = multiplying top and bottom by $c$
}}
{{eqn | r = \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = 
}}
{{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x \paren {\sqrt {a x^2 + b x + c} }^3} - \frac a c \int \frac {x^2 \rd x} {x \paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {x \rd x} {x \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = simplifying
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac a c \paren {\frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\dfrac x {\paren {\sqrt {a x^2 + b x + c} }^3}$
}}
{{eqn | o = 
      | ro= -
      | r = \frac b c \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac 1 c \paren {\frac {\paren {4 a b x + 2 b^2} - \paren {2 a b x + 4 a c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }
      | c = placing over a common denominator
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac 1 c \paren {\frac {2 a b x + 2 b^2 - 4 a c} {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }
      | c = simplifying
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac 1 c \paren {\frac {b^2 - 4 a c} {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } } + \frac 1 c \paren {\frac {2 a b x + b^2} {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }
      | c = splitting
}}
{{eqn | r = \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } }
      | c = rearranging and simplifying
}}
{{eqn | r = \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by Cube of Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} = -\frac {a x^2 + 2 b x + c} {c^2 x \sqrt {a x^2 + b x + c} } + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac {3 b} {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$

Proof:
{{begin-eqn}}
{{eqn | r = \int \frac {\d x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3}
      | o = 
      | c = 
}}
{{eqn | r = \int \frac {c \rd x} {c x^2 \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = multiplying top and bottom by $c$
}}
{{eqn | r = \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x^2 \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = 
}}
{{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} - \frac a c \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {x \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x^2 \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3}
      | c = simplifying
}}
{{eqn | r = \frac 1 c \paren {-\frac {\sqrt {a x^2 + b x + c} } {c x} - \frac b {2 c} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } } - \frac a c \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {a x^2 + b x + c} }$
}}
{{eqn | o = 
      | ro= -
      | r = \frac b c \paren {\frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} }
      | c = Primitive of $\dfrac 1 {x \paren {\sqrt {a x^2 + b x + c} }^3}$
}}
{{eqn | r = -\frac {a x^2 + b x + c} {c^2 x \sqrt {a x^2 + b x + c} } - \frac b {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac {2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = arranging common denominators for like terms
}}
{{eqn | o = 
      | ro= -
      | r = \frac {b x} {c^2 x \sqrt {a x^2 + b x + c} } - \frac b {c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac {b^2} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}
      | c = 
}}
{{eqn | r = -\frac {a x^2 + 2 b x + c} {c^2 x \sqrt {a x^2 + b x + c} } + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac {3 b} {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | c = assembling final form
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x by Half Integer Power of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x = \frac {\paren {a x^2 + b x + c}^{n + \frac 3 2} } {a \paren {2 n + 3} } - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
}}
{{eqn | r = \int \frac {2 a x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x} {2 a}
      | c = multiplying top and bottom by $2 a$
}}
{{eqn | r = \int \frac {\paren {2 a x + b - b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x} {2 a}
}}
{{eqn | n = 1
      | r = \frac 1 {2 a} \int \paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a x^2 + b x + c
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a x + b
      | c = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = \int \paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | r = \int z^{n + \frac 1 2} \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac {z^{n + \frac 3 2} } {n + \frac 3 2}
      | c = Primitive of Power
}}
{{eqn | r = \frac {2 z^{n + \frac 3 2} } {2 n + 3}
      | c = simplifying
}}
{{eqn | n = 2
      | r = \frac {2 \paren {a x^2 + b x + c}^{n + \frac 3 2} } {2 n + 3}
      | c = substituting for $z$
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | o = 
      | r = \int x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
}}
{{eqn | r = \frac 1 {2 a} \int \paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | c = from $(1)$
}}
{{eqn | r = \frac 1 {2 a} \paren {\frac {2 \paren {a x^2 + b x + c}^{n + \frac 3 2} } {2 n + 3} } - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | c = from $(2)$
}}
{{eqn | r = \frac {\paren {a x^2 + b x + c}^{n + \frac 3 2} } {a \paren {2 n + 3} } - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x cubed plus a cubed

:$\ds \int \frac {x \rd x} {x^3 + a^3} = \frac 1 {6 a} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {x^3 + a^3}
      | r = \int \frac {\paren {x + a - a} \rd x} {x^3 + a^3}
      | c = 
}}
{{eqn | r = \int \frac {\paren {x + a} \rd x} {x^3 + a^3} - a \int \frac {\d x} {x^3 + a^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int \frac {\paren {x + a} \rd x} {\paren {x + a} \paren {x^2 - a x + a^2} } - a \int \frac {\d x} {x^3 + a^3}
      | c = Sum of Two Cubes
}}
{{eqn | r = \int \frac {\d x} {x^2 - a x + a^2} - a \int \frac {\d x} {x^3 + a^3}
      | c = simplifying
}}
{{eqn | r = \frac 2 {a \sqrt 3} \arctan \paren {\frac {2 x - a} {a \sqrt 3} } - a \int \frac {\d x} {x^3 + a^3}
      | c = Primitive of $\dfrac 1 {x^2 - a x + a^2}$
}}
{{eqn | r = \frac 2 {a \sqrt 3} \, \map \arctan {\frac {2 x - a} {a \sqrt 3} } - a \paren {\frac 1 {6 a^2} \ln \size {\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 $\dfrac 1 {x^3 + a^3}$
}}
{{eqn | r = \frac 2 {a \sqrt 3} \, \map \arctan {\frac {2 x - a} {a \sqrt 3} } - \frac 1 {6 a} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } - \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}
      | c = simplifying
}}
{{eqn | r = \frac 1 {6 a} \, \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}
      | c = gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed plus a cubed

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 + a^3}
      | r = \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \frac {x - 2 a} {3 a^2 \paren {x^2 - a x + a^2} } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \frac {2 x - 4 a} {6 a^2 \paren {x^2 - a x + a^2} } } \rd x
      | c = multiplying top and bottom by $2$
}}
{{eqn | r = \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \paren {\frac {2 x - a} {6 a^2 \paren {x^2 - a x + a^2} } } - \paren {\frac {3 a} {6 a^2 \paren {x^2 - a x + a^2} } } } \rd x
      | c = 
}}
{{eqn | r = \frac 1 {3 a^2} \int \frac {\d x} {x + a} - \frac 1 {6 a^2} \int \frac {\paren {2 x - a} \rd x} {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {3 a^2} \ln \size {x + a} - \frac 1 {6 a^2} \int \frac {\paren {2 x - a} \rd x} {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | r = \frac 1 {3 a^2} \ln \size {x + a} - \frac 1 {6 a^2} \ln \size {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}
      | c = Primitive of Function under its Derivative
}}
{{eqn | r = \frac 1 {6 a^2} \ln \size {\paren {x + a}^2} - \frac 1 {6 a^2} \ln \size {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}
      | c = Logarithm of Power
}}
{{eqn | r = \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}
      | c = Difference of Logarithms
}}
{{eqn | r = \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {2 a} \paren {\frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} } }
      | c = Primitive of $\dfrac 1 {x^2 - a x + a^2}$
}}
{{eqn | r = \frac 1 {6 a^2} \ln \size {\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 = simplifying
}}
{{end-eqn}}
{{Qed}}
Theorem: Primitive of x squared over x cubed plus a cubed

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

Proof:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {x^3 + a^3}
      | r = 3 x^2
      | c = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x^2 \rd x} {x^3 + a^3}
      | r = \frac 1 3 \ln \size {x^3 + a^3} + C
      | c = Primitive of Function under its Derivative
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x cubed plus a cubed

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

Proof:
From Primitive of Power of x less one over Power of x plus Power of a:
:$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$
So:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {x^3 + a^3}
      | r = \frac 1 3 \ln \size {x^3 + a^3} + C
      | c = Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 3$
}}
{{end-eqn}}
directly.
{{qed}}
Theorem: Primitive of Reciprocal of x by x cubed plus a cubed

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

Proof:
From Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$:
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
Setting $n = 3$:
:$\ds \int \frac {\d x} {x \paren {x^3 + a^3} } = \frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } + C$
directly.
{{qed}}
Theorem: Primitive of Reciprocal of x squared by x cubed plus a cubed

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

Proof:
First a lemma: