hcju/proofwikiqa huggingface dataset:
First note that: {{begin-eqn}} {{eqn | l = x        | r = a \tan u       | c =  }} {{eqn | ll= \implies       | l = \sqrt {a^2 + x^2}       | r = \sqrt {a^2 + \left({a \tan u}\right)^2}       | c = }} {{eqn | r = a \sqrt {1 + \tan^2 u}       | c = taking $a$ outside the square root }} {{eqn | r = a \sqrt {\sec^2 u}       | c = Difference of Squares of Secant and Tangent }} {{eqn | n = 1       | r = a \sec u       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x        | r = a \tan u       | c =  }} {{eqn | ll= \implies       | l = \frac{\mathrm d x} {\mathrm d u}       | r = a \sec^2 u       | c = Derivative of Tangent Function }} {{eqn | ll= \implies       | l = \int F \left({\sqrt {a^2 + x^2} }\right) \ \mathrm d x       | r = \int a \sec^2 u \ F \left({\sqrt {a^2 + x^2} }\right) \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \int a \sec^2 u \ F \left({a \sec u}\right) \ \mathrm d u       | c = from $(1)$ }} {{eqn | r = a \int \sec^2 u \ F \left({a \sec u}\right) \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
First note that: {{begin-eqn}} {{eqn | l = x        | r = a \sec u       | c =  }} {{eqn | ll= \implies       | l = \sqrt {x^2 - a^2}       | r = \sqrt {\left({a \sec u}\right)^2 - a^2}       | c = }} {{eqn | r = a \sqrt {\sec^2 u - 1}       | c = taking $a$ outside the square root }} {{eqn | r = a \sqrt {\tan^2 u}       | c = Difference of Squares of Secant and Tangent }} {{eqn | n = 1       | r = a \tan u       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x        | r = a \sec u       | c =  }} {{eqn | ll= \implies       | l = \frac{\mathrm d x} {\mathrm d u}       | r = a \sec u \tan u       | c = Derivative of Secant Function }} {{eqn | ll= \implies       | l = \int F \left({\sqrt {x^2 - a^2} }\right) \ \mathrm d x       | r = \int a \sec u \tan u \ F \left({\sqrt {x^2 - a^2} }\right) \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \int a \sec u \tan u \ F \left({a \tan u}\right) \ \mathrm d u       | c = from $(1)$ }} {{eqn | r = a \int \sec u \tan u \ F \left({a \tan u}\right) \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = u        | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = a e^{a x}       | c = Derivative of Exponential Function: Corollary 1 }} {{eqn | r = a u       | c = Definition of $u$ }} {{eqn | ll= \leadsto       | l = \int \map F {e^{a x} } \rd x       | r = \int \frac {\map F u} {a u} \rd u       | c = Primitive of Composite Function }} {{eqn | r = \frac 1 a \int \frac {\map F u} u \rd u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = u        | r = \ln x       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac 1 x       | c = Derivative of Natural Logarithm Function }} {{eqn | ll= \implies       | l = \int F \left({\ln x}\right) \rd x       | r = \int F \left({u}\right) \ x \rd u       | c = Primitive of Composite Function }} {{eqn | r = \int F \left({u}\right) \ e^u \rd u       | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}} }} {{end-eqn}} {{qed}}
First note that: {{begin-eqn}} {{eqn | l = u       | r = \arcsin \frac x a       | c =  }} {{eqn | ll= \implies       | l = x       | r = a \sin u       | c = Definition of Arcsine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = u        | r = \arcsin \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac{\mathrm d u} {\mathrm d x}       | r = \frac 1 {\sqrt {a^2 - x^2} }       | c = Derivative of Arcsine Function: Corollary }} {{eqn | ll= \implies       | l = \int F \left({\arcsin \frac x a}\right) \ \mathrm d x       | r = \int F \left({u}\right) \ \sqrt {a^2 - x^2} \ \mathrm d u       | c = Primitive of Composite Function }} {{eqn | r = \int F \left({u}\right) \ \sqrt {a^2 - a^2 \sin^2 u} \ \mathrm d u       | c = Definition of $x$ }} {{eqn | r = \int F \left({u}\right) \ a \sqrt {1 - \sin^2 u} \ \mathrm d u       | c =  }} {{eqn | r = \int F \left({u}\right) \ a \cos u \ \mathrm d u       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = a \int F \left({u}\right) \ \cos u \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
First note that: {{begin-eqn}} {{eqn | l = u       | r = \arccos \frac x a       | c =  }} {{eqn | ll= \implies       | l = x       | r = a \cos u       | c = Definition of Arccosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = u        | r = \arccos \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac {-1} {\sqrt {a^2 - x^2} }       | c = Derivative of Arccosine Function: Corollary }} {{eqn | ll= \implies       | l = \int F \paren {\arccos \frac x a} \rd x       | r = \int F \paren u \ \paren {-\sqrt {a^2 - x^2} } \rd u       | c = Primitive of Composite Function }} {{eqn | r = \int F \paren u \ \paren {-\sqrt {a^2 - a^2 \cos^2 u} } \rd u       | c = Definition of $x$ }} {{eqn | r = \int F \paren u \ a \paren {-\sqrt {1 - \cos^2 u} } \rd u       | c =  }} {{eqn | r = \int F \paren u \ a \paren {-\sin u} \rd u       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = -a \int F \paren u \ \sin u \rd u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
First note that: {{begin-eqn}} {{eqn | l = u       | r = \arctan \frac x a       | c =  }} {{eqn | ll= \implies       | l = x       | r = a \tan u       | c = Definition of Arctangent }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = u        | r = \arctan \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac{\mathrm d u} {\mathrm d x}       | r = \frac a {a^2 + x^2}       | c = Derivative of Arctangent Function: Corollary }} {{eqn | ll= \implies       | l = \int F \left({\arctan \frac x a}\right) \ \mathrm d x       | r = \int F \left({u}\right) \ \frac {a^2 + x^2} a \ \mathrm d u       | c = Primitive of Composite Function }} {{eqn | r = \int F \left({u}\right) \ \frac {a^2 + a^2 \tan^2 u} a \ \mathrm d u       | c = Definition of $x$ }} {{eqn | r = \int F \left({u}\right) \ a^2 \frac {1 + \tan^2 u} a \ \mathrm d u       | c =  }} {{eqn | r = \int F \left({u}\right) \ a \sec^2 u \ \mathrm d u       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = a \int F \left({u}\right) \ \sec^2 u \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
First note that: {{begin-eqn}} {{eqn | l = u       | r = \operatorname{arccot} \frac x a       | c =  }} {{eqn | ll= \implies       | l = x       | r = a \cot u       | c = Definition of Arccotangent }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = u        | r = \operatorname{arccot} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac{\mathrm d u} {\mathrm d x}       | r = \frac {-a} {a^2 + x^2}       | c = Derivative of Arccotangent Function: Corollary }} {{eqn | ll= \implies       | l = \int F \left({\operatorname{arccot} \frac x a}\right) \ \mathrm d x       | r = \int F \left({u}\right) \ \frac {a^2 + x^2} {-a} \ \mathrm d u       | c = Primitive of Composite Function }} {{eqn | r = \int F \left({u}\right) \ \frac {a^2 + a^2 \cot^2 u} {-a} \ \mathrm d u       | c = Definition of $x$ }} {{eqn | r = \int F \left({u}\right) \ a^2 \frac {1 + \cot^2 u} {-a} \ \mathrm d u       | c =  }} {{eqn | r = \int F \left({u}\right) \ \left({-a}\right) \csc^2 u \ \mathrm d u       | c = Difference of Squares of Cosecant and Cotangent }} {{eqn | r = -a \int F \left({u}\right) \ \csc^2 u \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
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}}
First note that: {{begin-eqn}} {{eqn | l = u       | r = \operatorname{arccsc} \frac x a       | c =  }} {{eqn | ll= \implies       | l = x       | r = a \csc u       | c = Definition of Arccosecant }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = u        | r = \operatorname{arccsc} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac{\mathrm d u} {\mathrm d x}       | r = \frac {-a} {\left\vert{x}\right\vert {\sqrt {x^2 - a^2} } }       | c = Derivative of Arccosecant Function: Corollary }} {{eqn | ll= \implies       | l = \int F \left({\operatorname{arccsc} \frac x a}\right) \ \mathrm d x       | r = \int F \left({u}\right) \ \frac {\left\vert{x}\right\vert {\sqrt {x^2 - a^2} } } {-a} \ \mathrm d u       | c = Primitive of Composite Function }} {{eqn | r = \int F \left({u}\right) \ \frac {\left\vert{a \csc u}\right\vert {\sqrt {a^2 \csc^2 u - a^2} } } {-a} \ \mathrm d u       | c = Definition of $x$ }} {{eqn | r = \int F \left({u}\right) \ \left\vert{\csc u}\right\vert \left({-\sqrt {a^2 \csc^2 u - a^2} }\right) \ \mathrm d u       | c =  }} {{eqn | r = \int F \left({u}\right) \ \left({-a}\right) \left\vert{\csc u}\right\vert {\sqrt {\csc^2 u - 1} } \ \mathrm d u       | c =  }} {{eqn | r = \int F \left({u}\right) \ \left({-a}\right) \left\vert{\csc u}\right\vert \cot u \ \mathrm d u       | c = Difference of Squares of Cosecant and Cotangent }} {{eqn | r = -a \int F \left({u}\right) \ \left\vert{\csc u}\right\vert \cot u \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}} {{finish|Review the signs of the trig functions and their inverses rigorously -- I'm not sure about them.}}
{{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}}
Put $u = a x + b$. Then: {{begin-eqn}} {{eqn | l = x       | r = \frac {u - b} a       | c =  }} {{eqn | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 a       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3}       | r = \int \frac 1 a \frac {u - b} {a u^3} \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d u} {u^2} - \frac b {a^2} \int \frac {\mathrm d u} {u^3}       | c = Linear Combination of Integrals }} {{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 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C       | c = substituting for $u$ and rearranging }} {{end-eqn}} {{qed}}
From Primitive of $x$ by Power of $a x + b$: :$\displaystyle \int x \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^2} - \frac {b \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^2} + C$ where $n \ne - 1$ and $n \ne - 2$. The result follows by setting $n = -3$. {{qed}}
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 Integrals }} {{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}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$: :$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$ 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 Reciprocal of $a x + b$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {a x + b}       | r = \int \frac 1 a \frac {a x \ \mathrm d x} {a x + b}       | c =  }} {{eqn | r = \int \frac 1 a \frac {\left({a x + b - b}\right) \ \mathrm d x} {a x + b}       | c =  }} {{eqn | r = \frac 1 a \int \frac {\left({a x + b}\right) \ \mathrm d x} {a x + b} - \frac b a \int \frac {\mathrm d x} {a x + b}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 a \int \mathrm d x - \frac b a \int \frac {\mathrm d x} {a x + b}       | c = simplification }} {{eqn | r = \frac x a - \frac b a \int \frac {\mathrm d x} {a x + b}       | c = Primitive of Constant }} {{eqn | r = \frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{end-eqn}} {{qed}}
Put $u = a x + b$. Then: {{begin-eqn}} {{eqn | l = x       | r = \frac {u - b} a       | c =  }} {{eqn | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 a       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}       | r = \int \frac 1 a \frac {u - b} {a u^2} \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d u} u - \frac b {a^2} \int \frac {\mathrm d u} {u^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \ln \left\vert{u}\right\vert + C - \frac b {a^2} \int \frac {\mathrm d u} u       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^2} \ln \left\vert{u}\right\vert - \frac b {a^2} \frac {-1} u + C       | c = Primitive of Power }} {{eqn | r = \frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \ln \left\vert{a x + b}\right\vert + C       | c = substituting for $u$ and rearranging }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}       | r = \int \frac {a x \ \mathrm d x} {a \left({a x + b}\right)^2}       | c = multiplying top and bottom by $a$ }} {{eqn | r = \int \frac {\left({a x + b - b}\right) \ \mathrm d x} {a \left({a x + b}\right)^2}       | c = adding and subtracting $b$ }} {{eqn | r = \frac 1 a \int \frac {\left({a x + b}\right) \ \mathrm d x} {\left({a x + b}\right)^2} - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 a \int \frac {\mathrm d x} {a x + b} - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2}       | c = simplification }} {{eqn | r = \frac 1 a \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C       | c = Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)}$ }} {{eqn | r = \frac 1 a \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \left({-\frac 1 {a \left({a x + b}\right)} }\right) + C       | c = Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)^2}$ }} {{eqn | r = \frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \ln \left\vert{a x + b}\right\vert + C       | c = simplification }} {{end-eqn}} {{qed}}
{{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}}
{{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: :$\displaystyle \int_0^A e^{-s t} \map {f'} t \rd t$ By hypothesis, $f'$ is piecewise continuous with one-sided limits. So by Piecewise Continuous Function with One-Sided Limits is Darboux Integrable, this integral exists. This means that integration by parts can be invoked: :$\displaystyle \int h j\,' \rd t = h j - \int h' j \rd t$ 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: :$\displaystyle \lim_{A \mathop \to +\infty} e^{-s A} \map f A = 0$ which produces: :$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$ {{qed}}
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 Integrals }} {{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}}
From Reduction Formula for Primitive of $x^m \paren {a x + b}^n$: Decrement of Power of $x$: :$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$ 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}}
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 Integrals }} {{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}}