avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Function of Arctangent

:$\ds \int \map F {\arctan \frac x a} \rd x = a \int \map F u \sec^2 u \rd u$
where $u = \arctan \dfrac x a$.

Proof:
First note that:
{{begin-eqn}}
{{eqn | l = u
      | r = \arctan \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \tan u
      | c = {{Defof|Real Arctangent}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = u 
      | r = \arctan \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac a {a^2 + x^2}
      | c = Derivative of Arctangent Function: Corollary
}}
{{eqn | ll= \leadsto
      | l = \int \map F {\arctan \frac x a} \rd x
      | r = \int \map F u \frac {a^2 + x^2} a \rd u
      | c = Primitive of Composite Function
}}
{{eqn | r = \int \map F u \frac {a^2 + a^2 \tan^2 u} a \rd u
      | c = Definition of $x$
}}
{{eqn | r = \int \map F u a^2 \frac {1 + \tan^2 u} a \rd u
      | c = 
}}
{{eqn | r = \int \map F u a \sec^2 u \rd u
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = a \int \map F u \sec^2 u \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Function of Arccotangent

:$\ds \int \map F {\arccot \frac x a} \rd x = -a \int \map F u \csc^2 u \rd u$
where $u = \arccot \dfrac x a$.

Proof:
First note that:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccot \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \cot u
      | c = {{Defof|Arccotangent}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = u 
      | r = \arccot \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {-a} {a^2 + x^2}
      | c = Derivative of Arccotangent Function: Corollary
}}
{{eqn | ll= \leadsto
      | l = \int \map F {\arccot \frac x a} \rd x
      | r = \int \map F u \frac {a^2 + x^2} {-a} \rd u
      | c = Primitive of Composite Function
}}
{{eqn | r = \int \map F u \frac {a^2 + a^2 \cot^2 u} {-a} \rd u
      | c = Definition of $x$
}}
{{eqn | r = \int \map F u a^2 \frac {1 + \cot^2 u} {-a} \rd u
      | c = 
}}
{{eqn | r = \int \map F u \paren {-a} \csc^2 u \rd u
      | c = Difference of Squares of Cosecant and Cotangent
}}
{{eqn | r = -a \int \map F u \csc^2 u \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Function of Arcsecant

:$\ds \int \map F {\arcsec \frac x a} \rd x = a \int \map F u \sec u \tan u \rd u$
where $u = \arcsec \dfrac x a$.

Proof:
First note that:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \sec u
      | c = {{Defof|Arcsecant}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = u 
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac a {\size x {\sqrt {x^2 - a^2} } }
      | c = Derivative of Arcsecant Function: Corollary 1
}}
{{eqn | ll= \leadsto
      | l = \int \map F {\arcsec \frac x a} \rd x
      | r = \int \map F u \, \frac {\size x {\sqrt {x^2 - a^2} } } a \rd u
      | c = Primitive of Composite Function
}}
{{eqn | r = \int \map F u \frac {\size {a \sec u} {\sqrt {a^2 \sec^2 u - a^2} } } a \rd u
      | c = Definition of $x$
}}
{{eqn | r = \int \map F u \size {\sec u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u
      | c = 
}}
{{eqn | r = \int \map F u a \sec u {\sqrt {\sec^2 u - 1} } \rd u
      | c = $\sec u > 0$ in this domain
}}
{{eqn | r = \int \map F u a \sec u \tan u \rd u
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = a \int \map F u \sec u \tan u \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Function of Arccosecant

:$\ds \int \map F {\arccsc \frac x a} \rd x = -a \int \map F u \size {\csc u} \cot u \rd u$
where $u = \arccsc \dfrac x a$.

Proof:
First note that:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccsc \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \csc u
      | c = {{Defof|Arccosecant}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = u 
      | r = \arccsc \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {-a} {\size x {\sqrt {x^2 - a^2} } }
      | c = Derivative of Arccosecant Function: Corollary
}}
{{eqn | ll= \leadsto
      | l = \int \map F {\arccsc \frac x a} \rd x
      | r = \int \map F u \frac {\size x {\sqrt {x^2 - a^2} } } {-a} \rd u
      | c = Primitive of Composite Function
}}
{{eqn | r = \int \map F u \frac {\size {a \csc u} {\sqrt {a^2 \csc^2 u - a^2} } } {-a} \rd u
      | c = Definition of $x$
}}
{{eqn | r = \int \map F u \size {\csc u} \paren {-\sqrt {a^2 \csc^2 u - a^2} } \rd u
      | c = 
}}
{{eqn | r = \int \map F u \paren {-a} \size {\csc u} {\sqrt {\csc^2 u - 1} } \rd u
      | c = 
}}
{{eqn | r = \int \map F u \paren {-a} \size {\csc u} \cot u \rd u
      | c = Difference of Squares of Cosecant and Cotangent
}}
{{eqn | r = -a \int \map F u \size {\csc u} \cot u \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x + b

:$\ds \int \frac {\d x} {a x + b} = \frac 1 a \ln \size {a x + b} + C$
where $a \ne 0$ and $x \ne - \dfrac b a$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x + b}
      | r = \frac 1 a \int \frac {\map \d {a x + b} } {a x + b}
      | c = Primitive of Function of $a x + b$
}}
{{eqn | r = \frac 1 a \ln \size {a x + b} + C
      | c = Primitive of Reciprocal
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x + b

:$\ds \int \frac {\d x} {a x + b} = \frac 1 a \ln \size {a x + b} + C$
where $a \ne 0$ and $x \ne - \dfrac b a$.

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

:$\ds \int \frac {\d x} {a x + b} = \frac 1 a \ln \size {a x + b} + C$
where $a \ne 0$ and $x \ne - \dfrac b a$.

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 Reciprocal of a x + b

:$\ds \int \frac {\d x} {a x + b} = \frac 1 a \ln \size {a x + b} + C$
where $a \ne 0$ and $x \ne - \dfrac b a$.

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

:$\ds \int \frac {\d x} {a x + b} = \frac 1 a \ln \size {a x + b} + C$
where $a \ne 0$ and $x \ne - \dfrac b a$.

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

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

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}^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

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

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 over a x + b

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

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

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

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 over a x + b

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

Proof:
Put $u = a x + b$
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac {u - b} a
      | c = 
}}
{{eqn | l = \frac {\d x} {\d u}
      | r = \frac 1 a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {a x + b}
      | r = \int \frac 1 a \frac {u - b} {a u} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a^2} \int \rd u - \frac b {a^2} \int \frac {\d u} u
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac u {a^2} + C - \frac b {a^2} \int \frac {\d u} u
      | c = Primitive of Constant
}}
{{eqn | r = \frac u {a^2} - \frac b {a^2} \ln \size u + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {a x + b} {a^2} - \frac b {a^2} \ln \size {a x + b} + C
      | c = substituting for $u$
}}
{{eqn | r = \frac x a + \frac b {a^2} - \frac b {a^2} \ln \size {a x + b} + C
      | c = simplifying
}}
{{eqn | r = \frac x a - \frac b {a^2} \ln \size {a x + b} + C
      | c = subsuming $\dfrac b {a^2}$ into the arbitrary constant $C$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over a x + b

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

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: 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 $m = 1$ and $n = -1$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {a x + b}
      | r = \int x^1 \paren {a x + b}^{-1} \rd x
      | c = 
}}
{{eqn | r = \frac {x^1 \paren {a x + b}^0} {\paren 1 a} - \frac {1 b} {\paren 1 a} \int x^0 \paren {a x + b}^{-1} \rd x
      | c = 
}}
{{eqn | r = \frac x a - \frac b a \int \frac {\d x} {a x + b}
      | c = simplifying
}}
{{eqn | r = \frac x a - \frac b {a^2} \ln \size {a x + b} + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over a x + b

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {a x + b}
      | r = \int \frac 1 a \frac {a x \rd x} {a x + b}
      | c = 
}}
{{eqn | r = \int \frac 1 a \frac {\paren {a x + b - b} \rd x} {a x + b}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {\paren {a x + b} \rd x} {a x + b} - \frac b a \int \frac {\d x} {a x + b}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 a \int \d x - \frac b a \int \frac {\d x} {a x + b}
      | c = simplification
}}
{{eqn | r = \frac x a - \frac b a \int \frac {\d x} {a x + b}
      | c = Primitive of Constant
}}
{{eqn | r = \frac x a - \frac b {a^2} \ln \size {a x + b} + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{end-eqn}}
{{qed}}
Theorem: First Translation Property of Laplace Transforms

:$\laptrans {e^{a t} \map f t} = \map F {s - a}$
everywhere that $\laptrans f$ exists, for $\map \Re s > a$

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {e^{a t} \map f t}
      | r = \int_0^{\to +\infty} e^{-s t} e^{a t} \map f t \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \int_0^{\to +\infty} e^{-s t + a t} \map f t \rd t
      | c = Exponent Combination Laws
}}
{{eqn | r = \int_0^{\to +\infty} e^{-\paren {s - a} t} \map f t \rd t
}}
{{eqn | r = \map F {s - a}
      | c = {{Defof|Laplace Transform}}
}}
{{end-eqn}}
{{qed}}
Theorem: First Translation Property of Laplace Transforms

:$\laptrans {e^{a t} \map f t} = \map F {s - a}$
everywhere that $\laptrans f$ exists, for $\map \Re s > a$

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\cosh 5 t}
      | r = \dfrac 5 {s^2 - 5^2}
      | c = Laplace Transform of Hyperbolic Cosine
}}
{{eqn | ll= \leadsto
      | l = \laptrans {e^{4 t} \cosh 5 t}
      | r = \dfrac {s - 4} {\paren {s - 4}^2 - 25}
      | c = First Translation Property of Laplace Transforms
}}
{{eqn | r = \dfrac {s - 4} {s^2 - 8 s - 9}
      | c = multiplying out
}}
{{end-eqn}}
{{qed}}
Theorem: First Translation Property of Laplace Transforms

:$\laptrans {e^{a t} \map f t} = \map F {s - a}$
everywhere that $\laptrans f$ exists, for $\map \Re s > a$

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {e^{4 t} \cosh 5 t}
      | r = \laptrans {e^{4 t} \paren {\dfrac {e^{5 t} + e^{-5 t} } 2} }
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \dfrac 1 2 \laptrans {e^{9 t} + e^{-t} }
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {\dfrac 1 {s - 9} + \dfrac 1 {s + 1} }
      | c = Laplace Transform of Exponential
}}
{{eqn | r = \dfrac {s - 4} {s^2 - 8 s - 9}
      | c = mulitplying out
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Derivative

Let $f'$ be piecewise continuous with one-sided limits on said intervals.
Then $\laptrans f$ exists for $\map \Re s > a$, and:
:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\map {f'} t}
      | r = \int_0^{\mathop \to +\infty} e^{-s t} \map {f'} t \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t
      | c = {{Defof|Improper Integral on Closed Interval Unbounded Above}}
}}
{{end-eqn}}
Consider:
:$\ds \int_0^A e^{-s t} \map {f'} t \rd t$
We have {{hypothesis}} that $f'$ is piecewise continuous with one-sided limits.
Hence by Piecewise Continuous Function with One-Sided Limits is Darboux Integrable, this integral exists.
This means that integration by parts can be invoked:
:$\ds \int h j\,' \rd t = h j - \int h' j \rd t$
Here:
{{begin-eqn}}
{{eqn | l = h
      | r = e^{-s t}
}}
{{eqn | ll = \leadsto
      | l = h'
      | r = -s e^{-s t}
}}
{{eqn | l = j\,'
      | r = \map {f'} t
}}
{{eqn | ll = \leadsto
      | l = j
      | r = \map f t
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \int_0^A e^{-s t} \map {f'} t \rd t
      | r = \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = A} + s \int_0^A e^{-s t} \map f t \rd t
}}
{{end-eqn}}
Now, take the limit as $t = A \to +\infty$:
{{begin-eqn}}
{{eqn | l = \laptrans {\map {f'} t}
      | r = \lim_{A \mathop \to +\infty} e^{-s A} \map f A - \map f 0 + s \laptrans {\map f t}
}}
{{end-eqn}}
Recall that $f$ is of exponential order $a$:
{{begin-eqn}}
{{eqn | l = \size {\map f A}
      | o = <
      | r = K e^{a t}
}}
{{eqn | ll= \leadsto
      | l = \size {\map f A} \size {e^{-s A} }
      | o = <
      | r = K e^{a t} \size {e^{-s A} }
}}
{{eqn | ll= \leadsto
      | l = \size {e^{-s A} \map f A}
      | o = <
      | r = \size {K e^{a t} e^{-s A} }
      | c = Modulus of Product, Exponential Tends to Zero and Infinity
}}
{{eqn | r = \size {K e^{\paren {a - s} t} }
      | c = Exponent Combination Laws, definition of $A$
}}
{{eqn | r = \size {K e^{\paren {a - \map \Re s - i \map \Im s} t} }
}}
{{eqn | r = \size {K e^{\paren {a - \map \Re s} t} }
      | c = Modulus of Exponential is Modulus of Real Part
}}
{{eqn | r = K e^{\paren {a - \map \Re s} t} 
      | c = Exponential Tends to Zero and Infinity
}}
{{end-eqn}}
This implies, from Complex Exponential Tends to Zero and the Squeeze Theorem for Functions:
:$\ds \lim_{A \mathop \to +\infty} e^{-s A} \map f A = 0$
which produces:
:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$
{{qed}}
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:
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

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

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