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

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

Proof:
From Primitive of $x^m \paren {a x + b}^n$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Let $m = 2$ and $n = -1$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {a x + b}
      | r = \int x^2 \paren {a x + b}^{-1} \rd x
      | c = 
}}
{{eqn | r = \frac {x^2 \paren {a x + b}^0} {\paren 2 a} - \frac {2 b} {\paren 2 a} \int x^1 \paren {a x + b}^{-1} \rd x
      | c = 
}}
{{eqn | r = \frac {x^2} {2 a} - \frac b a \int \frac {x \rd x} {a x + b}
      | c = simplifying
}}
{{eqn | r = \frac {x^2} {2 a} - \frac b a \paren {\frac x a - \frac b {a^2} \ln \size {a x + b} } + C
      | c = Primitive of $x$ over $a x + b$
}}
{{eqn | r = \frac {x^2} {2 a} - \frac {b x} {a^2} + \frac {b^2} {a^3} \ln \size {a x + b} + C
      | c = simplifying
}}
{{eqn | r = \frac {a^2 x^2} {2 a^3} - \frac {2 a b x} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
      | c = 
}}
{{eqn | r = \frac {a^2 x^2 + 2 a b x} {2 a^3} - \frac {4 a b x} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
      | c = 
}}
{{eqn | r = \frac {a^2 x^2 + 2 a b x + b^2} {2 a^3} - \frac {4 a b x + b^2} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
      | c = 
}}
{{eqn | r = \frac {a^2 x^2 + 2 a b x + b^2} {2 a^3} - \frac {4 a b x + 4 b^2} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
      | c = subsuminging $\dfrac {3 b^2} {2 a^3}$ into the arbitrary constant
}}
{{eqn | r = \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 a b x + 2 b^2} {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
      | c = 
}}
{{eqn | r = \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 b \paren {a x + b} } {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over a x + b

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

Proof:
Put $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac {u - b} a
      | c = 
}}
{{eqn | l = \frac {\d u} {\d x}
      | r = \frac 1 a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^3 \rd x} {a x + b}
      | r = \int \frac 1 a \paren {\frac {u - b} a}^3 \frac {\d u} u
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\paren {u - b}^3} u \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 {a^4} \int \paren {u^2 - 3 b u + 3 b^2 - \frac {b^3} u} \rd u
      | c = multiplying out and simplifying
}}
{{eqn | r = \frac 1 {a^4} \paren {\int u^2 \rd u - \int 3 b u \rd u + \int 3 b^2 \rd u - \int \frac {b^3 \rd u} u}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + \int 3 b^2 \rd u - \int \frac {b^3} u \rd u} + C
      | c = Primitive of Power (two instances)
}}
{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + b^2 u - \int \frac {b^3} u \rd u} + C
      | c = Primitive of Constant and subsuming arbitrary constant $C$
}}
{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + 3 b^2 u - b^3 \int \frac {\d u} u} + C
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + 3 b^2 u - b^3 \ln \size u} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {\paren {a x + b}^3} {3 a^4} - \frac {3 b \paren {a x + b}^2} {2 a^4} - \frac {3 b^2 \paren {a x + b} } {a^4} + \frac {b^3} {a^4} \ln \size {a x + b} + C
      | c = substituting for $u$ and simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by a x + b

:$\ds \int \frac {\rd x} {x \paren {a x + b} } = \frac 1 b \ln \size {\frac x {a x + b} } + C$

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

:$\ds \int \frac {\rd x} {x \paren {a x + b} } = \frac 1 b \ln \size {\frac x {a x + b} } + C$

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

:$\ds \int \frac {\rd x} {x \paren {a x + b} } = \frac 1 b \ln \size {\frac x {a x + b} } + C$

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

:$\ds \int \frac {\rd x} {x \paren {a x + b} } = \frac 1 b \ln \size {\frac x {a x + b} } + C$

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

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

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

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

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

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

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

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b} }
      | r = \int \frac {b \rd x} {b x^3 \paren {a x + b} }
      | c = multiplying top and bottom by $b$
}}
{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x^3 \paren {a x + b} }
      | c = adding and subtracting $a x$
}}
{{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x^3 \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x^3 \paren {a x + b} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 b \int \frac {\d x} {x^3} - \frac a b \int \frac {\d x} {x^2 \paren {a x + b} }
      | c = simplification
}}
{{eqn | r = -\frac 1 {2 b x^2} - \frac a b \int \frac {\d x} {x^2 \paren {a x + b} } + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac 1 {2 b x^2} - \frac a b \paren {-\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} } + C
      | c = Primitive of $\dfrac 1 {x^2 \paren {a x + b} }$
}}
{{eqn | r = -\frac 1 {2 b x^2} + \frac a {b^2 x} - \frac {a^2} {b^3} \ln \size {\frac {a x + b} x} + C
      | c = multiplying out
}}
{{eqn | r = \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C
      | c = Difference of Logarithms
}}
{{eqn | r = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x + b squared

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

Proof:
Let $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a x + b}^2}
      | r = \frac 1 a \int \frac {\d u} {u^2}
      | c = Primitive of Function of $a x + b$
}}
{{eqn | r = \frac 1 a \frac {-1} u + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac 1 {a \paren {a x + b} } + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x + b squared

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

Proof:
From Primitive of Power of $a x + b$:
:$\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$
where $n \ne 1$.
The result follows by setting $n = -2$.
{{qed}}
Theorem: Primitive of x over a x + b squared

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

Proof:
Put $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac {u - b} a
      | c = 
}}
{{eqn | l = \frac {\d u} {\d x}
      | r = \frac 1 a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {\paren {a x + b}^2}
      | r = \int \frac 1 a \frac {u - b} {a u^2} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d u} u - \frac b {a^2} \int \frac {\d u} {u^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \ln \size u + C - \frac b {a^2} \int \frac {\d u} u
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^2} \ln \size u - \frac b {a^2} \frac {-1} u + C
      | c = Primitive of Power
}}
{{eqn | r = \frac b {a^2 \paren {a x + b} } + \frac 1 {a^2} \ln \size {a x + b} + C
      | c = substituting for $u$ and rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over a x + b squared

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {\paren {a x + b}^2}
      | r = \int \frac {a x \rd x} {a \paren {a x + b}^2}
      | c = multiplying top and bottom by $a$
}}
{{eqn | r = \int \frac {\paren {a x + b - b} \rd x} {a \paren {a x + b}^2}
      | c = adding and subtracting $b$
}}
{{eqn | r = \frac 1 a \int \frac {\paren {a x + b} \rd x} {\paren {a x + b}^2} - \frac b a \int \frac {\d x} {\paren {a x + b}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 a \int \frac {\d x} {a x + b} - \frac b a \int \frac {\d x} {\paren {a x + b}^2}
      | c = simplification
}}
{{eqn | r = \frac 1 a \paren {\frac 1 a \ln \size {a x + b} } - \frac b a \int \frac {\d x} {\paren {a x + b}^2} + C
      | c = Primitive of Reciprocal of $\dfrac 1 {a x + b}$
}}
{{eqn | r = \frac 1 a \paren {\frac 1 a \ln \size {a x + b} } - \frac b a \paren {-\frac 1 {a \paren {a x + b} } } + C
      | c = Primitive of Reciprocal of $\dfrac 1 {\paren {a x + b}^2}$
}}
{{eqn | r = \frac b {a^2 \paren {a x + b} } + \frac 1 {a^2} \ln \size {a x + b} + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over a x + b squared

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

Proof:
Put $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac {u - b} a
      | c = 
}}
{{eqn | l = \frac {\d u} {\d x}
      | r = \frac 1 a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\paren {a x + b}^2}
      | r = \int \frac 1 a \paren {\frac {u - b} a}^2 \frac 1 {u^2} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac 1 {a^3} \paren {1 - \frac {2 b} u + \frac {b^2} {u^2} } \rd u
      | c = Square of Difference, and simplification
}}
{{eqn | r = \frac 1 {a^3} \int \d u - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C
      | c = Primitive of Constant
}}
{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \frac {-1} u + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C
      | c = substituting for $u$ and rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over a x + b squared

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\paren {a x + b}^2}
      | r = \int \frac {a x^2 \rd x} {a \paren {a x + b}^2}
      | c = multiplying top and bottom by $a$
}}
{{eqn | r = \int \frac {x \paren {a x + b - b} \rd x} {a \paren {a x + b}^2}
      | c = adding and subtracting $b x$
}}
{{eqn | r = \frac 1 a \int \frac {x \paren {a x + b} \rd x} {\paren {a x + b}^2} - \frac b a \int \frac {x \rd x} {\paren {a x + b}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 a \int \frac {x \rd x} {a x + b} - \frac b a \int \frac {x \rd x} {\paren {a x + b}^2}
      | c = simplification
}}
{{eqn | r = \frac 1 a \paren {\frac x a - \frac b {a^2} \ln \size {a x + b} } - \frac b a \int \frac {\d x} {\paren {a x + b}^2} + C
      | c = Primitive of Reciprocal of $\dfrac x {\paren {a x + b} }$
}}
{{eqn | r = \frac 1 a \paren {\frac x a - \frac b {a^2} \ln \size {a x + b} } - \frac b a \paren {\frac b {a^2 \paren {a x + b} } + \frac 1 {a^2} \ln \size {a x + b} } + C
      | c = Primitive of Reciprocal of $\dfrac x {\paren {a x + b}^2}$
}}
{{eqn | r = \frac x {a^2} - \frac b {a^3} \ln \size {a x + b} - \frac {b^2} {a^3 \paren {a x + b} } - \frac b {a^3} \ln \size {a x + b} + C
      | c = multiplying out
}}
{{eqn | r = \frac x {a^2} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C
      | c = gathering terms
}}
{{eqn | r = \frac {a x} {a^3} + \frac b {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C
      | c = where $\dfrac b {a^3}$ is subsumed into $C$
}}
{{eqn | r = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over a x + b squared

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

Proof:
Put $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac {u - b} a
      | c = 
}}
{{eqn | l = \frac {\d u} {\d x}
      | r = \frac 1 a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l =  \int \frac {x^3 \rd x} {\paren {a x + b}^2}
      | r = \int \frac 1 a \paren {\frac {u - b} a}^3 \frac 1 {u^2} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac 1 {a^4} \paren {u - 3 b + \frac {3 b^2} u - \frac {b^3} {u^2} } \rd u
      | c = multiplying out and simplification
}}
{{eqn | r = \frac 1 {a^4} \int u \rd u - \frac {3 b} {a^4} \int \d u + \frac {3 b^2} {a^4} \int \frac {\d u} u - \frac {b^3} {a^4} \int \frac {\d u} {u^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {u^2} {2 a^4} - \frac {3 b} {a^4} \int \d u + \frac {3 b^2} {a^4} \int \frac {\d u} u - \frac {b^3} {a^4} \frac {-1} u + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {u^2} {2 a^4} - \frac {3 b u} {a^4} + \frac {3 b^2} {a^4} \int \frac {\d u} u + \frac {b^3} {a^4 u} + C
      | c = Primitive of Constant
}}
{{eqn | r = \frac {u^2} {2 a^4} - \frac {3 b u} {a^4} + \frac {3 b^2} {a^4} \ln \size u + \frac {b^3} {a^4 u} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {\paren {a x + b}^2} {2 a^4} -  \frac {3 b \paren {a x + b} } {a^4} + \frac {b^3} {a^4 \paren {a x + b} } + \frac {3 b^2} {a^4} \ln \size {a x + b} + C
      | c = substituting for $u$ and rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by a x + b squared

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

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