avewright/proofwiki-math huggingface dataset:
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 \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 squared by a x + b squared

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

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

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

Proof:
Let $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\rd x} {\paren {a x + b}^3}
      | r = \frac 1 a \int \frac {\rd u} {u^3}
      | c = Primitive of Function of $a x + b$
}}
{{eqn | r = \frac 1 a \frac {-1} {2 u^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 a \paren {a x + b}^2} + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x + b cubed

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

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 $n = -3$.
{{qed}}
Theorem: Primitive of x over a x + b cubed

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

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}^3}
      | r = \int \frac 1 a \frac {u - b} {a u^3} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d u} {u^2} - \frac b {a^2} \int \frac {\d u} {u^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \frac {-1} u - \frac b {a^2} \frac {-1} {2 u^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {a^2 \paren {a x + b} } + \frac b {2 a^2 \paren {a x + b}^2} + C
      | c = substituting for $u$ and rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over a x + b cubed

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

Proof:
From Primitive of $x$ by Power of $a x + b$:
:$\ds \int x \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 2} } {\paren {n + 2} a^2} - \frac {b \paren {a x + b}^{n + 1} } {\paren {n + 1} a^2} + C$
where $n \ne - 1$ and $n \ne - 2$.
The result follows by setting $n = -3$.
{{qed}}
Theorem: Primitive of x squared over a x + b cubed

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

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}^3}
      | 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 {\frac 1 u - \frac {2 b} {u^2} + \frac {b^2} {u^3} } \rd u
      | c = Square of Difference, and simplification
}}
{{eqn | r = \frac 1 {a^3} \int \frac {\d u} u - \frac {2 b} {a^3} \int \frac {\d u} {u^2} + \frac {b^2} {a^3} \int \frac {\d u} {u^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^3} \ln \size u - \frac {2 b} {a^3} \int \frac {\d u} {u^2} + \frac {b^2} {a^3} \int \frac {\d u} {u^3} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^3} \ln \size u - \frac {2 b} {a^3} \frac {-1} u + \frac {b^2} {a^3} \frac {-1} {2 u^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {2 b} {a^3 \paren {a x + b} } - \frac {b^2} {2 a^3 \paren {a x + b}^2} + \frac 1 {a^3} \ln \size {a x + b} + C
      | c = substituting for $u$ and rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over a x + b cubed

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

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}^3}
      | r = \int \frac 1 a \paren {\frac {u - b} a}^3 \frac 1 {u^3} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac 1 {a^4} \paren {1 - \frac {3 b} u + \frac {3 b^2} {u^2} - \frac {b^3} {u^3} } \rd u
      | c = multiplying out and simplification
}}
{{eqn | r = \frac 1 {a^4} \int \d u - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \int \frac {\d u} {u^2} - \frac {b^3} {a^4} \int \frac {\d u} {u^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac u {a^4} - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \int \frac {\d u} {u^2} - \frac {b^3} {a^4} \int \frac {\d u} {u^3} + C
      | c = Primitive of Constant
}}
{{eqn | r = \frac u {a^4} - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \frac {-1} u - \frac {b^3} {a^4} \frac {-1} {2 u^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac u {a^4} - \frac {3 b} {a^4} \ln \size u - \frac {3 b^2} {a^4 u} + \frac {b^3} {2 a^4 u^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {a x + b} {a^4} - \frac {3 b^2} {a^4 \paren {a x + b} )} + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C
      | c = substituting for $u$ and rearranging
}}
{{eqn | r = \frac x {a^3} + \frac b {a^4} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C
      | c = separating first term
}}
{{eqn | r = \frac x {a^3} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C
      | c = subsuming $\dfrac b {a^4}$ into the arbitrary constant $C$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by a x + b cubed

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {a x + b}^3}
      | r = \int \paren {\frac {-3 a} {b^4 x} + \frac 1 {b^3 x^2} + \frac {3 a^2} {b^4 \paren {a x + b} } + \frac {2 a^2} {b^3 \paren {a x + b}\^2} + \frac {a^2} {b^2 \paren {a x + b}^3} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac {-3 a} {b^4} \int \frac {\d x} x + \frac 1 {b^3} \int \frac {\d x} {x^2} + \frac {3 a^2} {b^4} \int \frac {\d x} {\paren {a x + b} } + \frac {2 a^2} {b^3} \int \frac {\d x} {\paren {a x + b}^2} + \frac {a^2} {b^2} \int \frac {\d x} {\paren {a x + b}^3}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {-3 a} {b^4} \ln \size x + \frac 1 {b^3} \int \frac {\d x} {x^2} + \frac {3 a^2} {b^4} \int \frac {\d x} {\paren {a x + b} } + \frac {2 a^2} {b^3} \int \frac {\d x} {\paren {a x + b}^2} + \frac {a^2} {b^2} \int \frac {\d x} {\paren {a x + b}^3} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {-3 a} {b^4} \ln \size x + \frac 1 {b^3} \int \frac {\d x} {x^2} + \frac {3 a^2} {b^4} \ln \size {a x + b} + \frac {2 a^2} {b^3} \int \frac {\d x} {\paren {a x + b}^2} + \frac {a^2} {b^2} \int \frac {\d x} {\paren {a x + b}^3} + C
      | c = Primitive of Reciprocal of a x + b
}}
{{eqn | r = \frac {-3 a} {b^4} \ln \size x + \frac 1 {b^3} \int \frac {\d x} {x^2} + \frac {3 a^2} {b^4} \ln \size {a x + b} + \frac {2 a^2} {b^3} \frac {-1} {a \paren {a x + b} } + \frac {a^2} {b^2} \int \frac {\d x} {\paren {a x + b}^3} + C
      | c = Primitive of Reciprocal of a x + b squared
}}
{{eqn | r = \frac {-3 a} {b^4} \ln \size x + \frac 1 {b^3} \int \frac {\d x} {x^2} + \frac {3 a^2} {b^4} \ln \size {a x + b} - \frac {2 a} {b^3 \paren {a x + b} } + \frac {a^2} {b^2} \frac {-1} {2 a \paren {a x + b}^2} + C
      | c = Primitive of Reciprocal of a x + b cubed
}}
{{eqn | r = \frac {-3 a} {b^4} \ln \size x + \frac 1 {b^3} \frac {-1} x + \frac {3 a^2} {b^4} \ln \size {a x + b} - \frac {2 a} {b^3 \paren {a x + b} } - \frac a {2 b^2 \paren {a x + b}^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-a} {2 b^2 \paren {a x + b}^2} - \frac {2 a} {b^3 \paren {a x + b} } - \frac 1 {b^3 x} + \frac {3 a} {b^4} \ln \size {\frac {a x + b} x} + C
      | c = Difference of Logarithms and rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Second Derivative

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le a$.
Let $f$ be twice differentiable.
Let $f'$ be continuous and $f' '$ piecewise continuous with one-sided limits on said intervals.
Let $f$ and $f'$ be of exponential order.
Let $\laptrans f$ denote the Laplace transform of $f$.
Then $\laptrans {f' '}$ exists for $\map \Re s > a$, and:
:$\laptrans {\map {f' '} t} = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0$

Proof:
{{Explain|Why does $\laptrans {f' '}$ exist? It needs to be proven}}
{{begin-eqn}}
{{eqn | l = \laptrans {\map {f' '} t}
      | r = s \laptrans {\map {f'} t} - \map {f'} 0
      | c = Laplace Transform of Derivative
}}
{{eqn | r = s \paren {s \laptrans {\map f t} - \map f 0} - \map {f'} 0
      | c = Laplace Transform of Derivative
}}
{{eqn | r = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Number is Transcendental

Euler's Number $e$ is transcendental.

Proof:
{{ProofWanted}}
{{qed}}
Theorem: Laplace Transform of Higher Order Derivatives

{{begin-eqn}}
{{eqn | l = \laptrans {\map {f^{\paren n} } t}
      | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^n s^{j - 1} \map {f^{\paren {n - j} } } 0
}}
{{eqn | r = s^n \map F s - s^{n - 1} \, \map f 0 - s^{n - 2} \, \map {f'} 0 - s^{n - 3} \, \map {f' '} 0 - \ldots - s \, \map {f^{\paren {n - 2} } } 0 - \map {f^{\paren {n - 1} } } 0
}}
{{end-eqn}}

Proof:
The proof proceeds by induction on $n$, the order of the derivative of $f$.
Theorem: Laplace Transform of Hyperbolic Cosine

Let $\cosh t$ be the hyperbolic cosine, where $t$ is real.
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
Then:
:$\laptrans {\cosh a t} = \dfrac s {s^2 - a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\cosh at}
      | r = \laptrans {\frac {e^{at} + e^{-at} } 2}
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac 1 2 \paren {\laptrans {e^{at} } + \laptrans {e^{-at} } }
      | c = Linear Combination of Laplace Transforms
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 {s - a} + \frac 1 {s + a} }
      | c = Laplace Transform of Exponential
}}
{{eqn | r = \frac 1 2 \paren {\frac {s + a + s - a} {\paren {s - a} \paren {s + a} } }
}}
{{eqn | r = \frac s {s^2 - a^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Hyperbolic Cosine

Let $\cosh t$ be the hyperbolic cosine, where $t$ is real.
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
Then:
:$\laptrans {\cosh a t} = \dfrac s {s^2 - a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\cosh a t}
      | r = \laptrans {\frac {e^{a t} + e^{-a t} } 2}
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \int_0^{\to +\infty} e^{-s t} \paren {\frac {e^{a t} + e^{-a t} } 2} \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t + \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{-a t} \rd t
      | c = Linear Combination of Laplace Transforms
}}
{{eqn | r = \dfrac 1 2 \laptrans {e^{a t} } + \dfrac 1 2 \laptrans {e^{-a t} }
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 {s - a} + \frac 1 {s + a} }
      | c = Laplace Transform of Exponential
}}
{{eqn | r = \frac 1 2 \paren {\frac {s + a + s - a} {\paren {s - a} \paren {s + a} } }
}}
{{eqn | r = \frac s {s^2 - a^2}
}}
{{end-eqn}}
{{Qed}}
Theorem: Laplace Transform of Hyperbolic Sine

Let $\sinh t$ be the hyperbolic sine, where $t$ is real.
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
Then:
:$\laptrans {\sinh a t} = \dfrac a {s^2 - a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\sinh {a t} }
      | r = \int_0^{\to +\infty} e^{-s t} \sinh {a t} \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \intlimits {\frac {e^{-s t} \paren {-s \sinh a t - a \cosh a t} } {\paren {-s}^2 - a^2} } {t \mathop = 0} {t \mathop \to +\infty}
      | c = Primitive of $e^{a x} \sinh b x$
}}
{{eqn | r = 0 - \frac {-s \, \map \sinh {0 \times a} - a \, \map \cosh {0 \times a} } {s^2 - a^2}
      | c = Exponential Tends to Zero, Exponential of Zero
}}
{{eqn | r = \frac {s \sinh 0 + a \cosh 0} {s^2 - a^2}
      | c = simplifying
}}
{{eqn | r = \frac a {s^2 - a^2}
      | c = Hyperbolic Sine of Zero is Zero, Hyperbolic Cosine of Zero is One
}}
{{end-eqn}}
{{Qed}}
Theorem: Laplace Transform of Hyperbolic Sine

Let $\sinh t$ be the hyperbolic sine, where $t$ is real.
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
Then:
:$\laptrans {\sinh a t} = \dfrac a {s^2 - a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\sinh a t}
      | r = \laptrans {\frac {e^{a t} - e^{-a t} } 2}
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac 1 2 \paren {\laptrans {e^{at} } - \laptrans {e^{-a t} } }
      | c = Linear Combination of Laplace Transforms
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 {s-a} - \frac 1 {s + a} }
      | c = Laplace Transform of Exponential
}}
{{eqn | r = \frac 1 2 \paren {\frac {s + a - s + a} {\paren {s - a} \paren {s + a} } }
}}
{{eqn | r = \frac a {s^2 - a^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Hyperbolic Sine

Let $\sinh t$ be the hyperbolic sine, where $t$ is real.
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
Then:
:$\laptrans {\sinh a t} = \dfrac a {s^2 - a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\sinh a t}
      | r = \laptrans {\frac {e^{a t} - e^{-a t} } 2}
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \int_0^{\to +\infty} e^{-s t} \paren {\frac {e^{a t} - e^{-a t} } 2} \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t - \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{-a t} \rd t
      | c = Linear Combination of Laplace Transforms
}}
{{eqn | r = \dfrac 1 2 \laptrans {e^{a t} } - \dfrac 1 2 \laptrans {e^{-a t} }
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 {s-a} - \frac 1 {s + a} }
      | c = Laplace Transform of Exponential
}}
{{eqn | r = \frac 1 2 \paren {\frac {s + a - s + a} {\paren {s - a} \paren {s + a} } }
}}
{{eqn | r = \frac a {s^2 - a^2}
}}
{{end-eqn}}
{{Qed}}
Theorem: Primitive of Power of a x + b

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

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

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

Proof:
Let $u = a x + b$.
Then:
:$\dfrac {\d u} {\d x} = a$
Then:
{{begin-eqn}}
{{eqn | l = \int \paren {a x + b}^n \rd x
      | r = \int \dfrac {u^n} a \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \dfrac 1 a \dfrac {u^{n + 1} } {n + 1}
      | c = Primitive of Power
}}
{{eqn | r = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C
      | c = substituting back for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of a x + b

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

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} }
      | r = \dfrac {\paren {n + 1} \paren {a x + b}^n} {\paren {n + 1} a} \map {\dfrac \d {\d x} } {a x + b}
      | c = Power Rule for Derivatives, Chain Rule for Derivatives
}}
{{eqn | r = \dfrac {a \paren {n + 1} \paren {a x + b}^n} {\paren {n + 1} a}
      | c = Power Rule for Derivatives
}}
{{eqn | r = \paren {a x + b}^n
      | c = simplifying
}}
{{end-eqn}}
The result follows by definition of primitive.
{{qed}}
Theorem: Primitive of x by Power of a x + b

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

Proof:
Let $u = a x + b$.
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 x \paren {a x + b}^n \rd x
      | r = \frac 1 a \int \frac {u - b} a u^n \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a^2} \int u^{n + 1} \rd u - \frac b {a^2} \int u^n \rd u
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \frac {u^{n + 2} } {n + 2} - \frac b {a^2} \frac {u^{n + 1} } {n + 1} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {\paren {a x + b}^{n + 2} } {\paren {n + 2} a^2} - \frac {b \paren {a x + b}^{n + 1} } {\paren {n + 1} a^2} + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Power of a x + b

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

Proof:
Let $u = a x + b$.
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 x \paren {a x + b}^n \rd x
      | r = \int \frac 1 a \paren {\frac {u - b} a}^2 u^n \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac 1 {a^3} \paren {u^{n + 2} - 2 b u^{n + 1} + b^2 u^n} \rd u
      | c = Square of Difference and multiplying out
}}
{{eqn | r = \frac 1 {a^3} \int u^{n + 3} \rd u - \frac {2 b} {a^3} \int u^{n + 1} \rd u + \frac {b^2} {a^3} \int u^n \rd u
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^3} \frac {u^{n + 3} } {n + 3} - \frac {2 b} {a^3} \frac {u^{n + 2} } {n + 2} + \frac {b^2} {a^3} \frac {u^{n + 1} } {n + 1} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {\paren {a x + b}^{n + 3} } {\paren {n + 3} a^3} - \frac {2 b \paren {a x + b}^{n + 2} } {\paren {n + 2} a^3} + \frac {b^2 \paren {a x + b}^{n + 1} } {\paren {n + 1} a^3} + C
      | c = substituting for $u$
}}
{{end-eqn}}
{{qed}}