hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2}       | r = \int \frac {a x^2 \ \mathrm d x} {a \left({a x + b}\right)^2}       | c = multiplying top and bottom by $a$ }} {{eqn | r = \int \frac {x \left({a x + b - b}\right) \ \mathrm d x} {a \left({a x + b}\right)^2}       | c = adding and subtracting $b x$ }} {{eqn | r = \frac 1 a \int \frac {x \left({a x + b}\right) \ \mathrm d x} {\left({a x + b}\right)^2} - \frac b a \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 a \int \frac {x \ \mathrm d x} {a x + b} - \frac b a \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}       | c = simplification }} {{eqn | r = \frac 1 a \left({\frac x a - \frac b {a^2} \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 $\dfrac x {\left({a x + b}\right)}$ }} {{eqn | r = \frac 1 a \left({\frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \left({\frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \left\vert{a x + b}\right\vert }\right) + C       | c = Primitive of Reciprocal of $\dfrac x {\left({a x + b}\right)^2}$ }} {{eqn | r = \frac x {a^2} - \frac b {a^3} \ln \left\vert{a x + b}\right\vert - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac b {a^3} \ln \left\vert{a x + b}\right\vert + C       | c = multiplying out }} {{eqn | r = \frac x {a^2} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C       | c = gathering terms }} {{eqn | r = \frac {a x} {a^3} + \frac b {a^3} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C       | c = where $\dfrac b {a^3}$ is subsumed into $C$ }} {{eqn | r = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C       | c = simplification }} {{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^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 Integrals }} {{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}}
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^3 \ \mathrm d x} {\left({a x + b}\right)^2}       | r = \int \frac 1 a \left({\frac {u - b} a}\right)^3 \frac 1 {u^2} \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \int \frac 1 {a^4} \left({u - 3 b + \frac {3 b^2} u - \frac {b^3} {u^2} }\right) \ \mathrm d u       | c = multiplying out and simplification }} {{eqn | r = \frac 1 {a^4} \int u \ \mathrm d u - \frac {3 b} {a^4} \int \mathrm d u + \frac {3 b^2} {a^4} \int \frac {\mathrm d u} u - \frac {b^3} {a^4} \int \frac {\mathrm d u} {u^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac {u^2} {2 a^4} - \frac {3 b} {a^4} \int \mathrm d u + \frac {3 b^2} {a^4} \int \frac {\mathrm 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 {\mathrm 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 \left\vert{u}\right\vert + \frac {b^3} {a^4 u} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {\left({a x + b}\right)^2} {2 a^4} -  \frac {3 b \left({a x + b}\right)} {a^4} + \frac {b^3} {a^4 \left({a x + b}\right)} + \frac {3 b^2} {a^4} \ln \left\vert{a x + b}\right\vert + C       | c = substituting for $u$ and rearranging }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)^2}       | r = \int \left({-\frac {2 a} {b^3 x} + \frac 1 {b^2 x^2} + \frac {2 a^2} {b^3 \left({a x + b}\right)} + \frac {a^2} {b^2 \left({a x + b}\right)^2} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = -\frac {2 a} {b^3} \int \frac {\mathrm d x} x + \frac 1 {b^2} \int \frac {\mathrm d x} {x^2} + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = -\frac {2 a} {b^3} \int \frac {\mathrm d x} x + \frac 1 {b^2} \frac {-1} x + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C       | c = Primitive of Power }} {{eqn | r = -\frac {2 a} {b^3} \ln \left\vert{x}\right\vert + \frac 1 {b^2} \frac {-1} x + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C       | c = Primitive of $\dfrac 1 x$ }} {{eqn | r = -\frac {2 a} {b^3} \ln \left\vert{x}\right\vert - \frac 1 {b^2 x} + \frac {2 a^2} {a b^3} \ln \left\vert{a x + b}\right\vert + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | r = -\frac {2 a} {b^3} \ln \left\vert{x}\right\vert - \frac 1 {b^2 x} + \frac {2 a} {b^3} \ln \left\vert{a x + b}\right\vert + \frac {a^2} {b^2} \frac {-1} {a \left({a x + b}\right)} + C       | c = Primitive of $\dfrac 1 {\left({a x + b}\right)^2}$ }} {{eqn | r = \frac {-a} {b^2 \left({a x + b}\right)} - \frac 1 {b^2 x} + \frac {2 a} {b^3} \ln \left\vert{\frac {a x + b} x}\right\vert + C       | c = Difference of Logarithms and rearranging }} {{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}^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 Integrals }} {{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}}
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 Integrals }} {{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}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)^3}       | r = \int \left({\frac {-3 a} {b^4 x} + \frac 1 {b^3 x^2} + \frac {3 a^2} {b^4 \left({a x + b}\right)} + \frac {2 a^2} {b^3 \left({a x + b}\right)^2} + \frac {a^2} {b^2 \left({a x + b}\right)^3} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac {-3 a} {b^4} \int \frac {\mathrm d x} x + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3}       | c = Linear Combination of Integrals }} {{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C       | c = Primitive of Reciprocal of a x + b }} {{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert + \frac {2 a^2} {b^3} \frac {-1} {a \left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C       | c = Primitive of Reciprocal of a x + b squared }} {{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert - \frac {2 a} {b^3 \left({a x + b}\right)} + \frac {a^2} {b^2} \frac {-1} {2 a \left({a x + b}\right)^2} + C       | c = Primitive of Reciprocal of a x + b cubed }} {{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \frac {-1} x + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert - \frac {2 a} {b^3 \left({a x + b}\right)} - \frac a {2 b^2 \left({a x + b}\right)^2} + C       | c = Primitive of Power }} {{eqn | r = \frac {-a} {2 b^2 \left({a x + b}\right)^2} - \frac {2 a} {b^3 \left({a x + b}\right)} - \frac 1 {b^3 x} + \frac {3 a} {b^4} \ln \left\vert{\frac {a x + b} x}\right\vert + C       | c = Difference of Logarithms and rearranging }} {{end-eqn}} {{qed}}
{{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}}
The proof proceeds by induction on $n$, the order of the derivative of $f$. === Basis for the Induction === By hypothesis $f$ is of exponential order. Thus Laplace Transform of Derivative can be invoked: :$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$ This is the basis for the induction. === Induction Hypothesis === Fix $n - 1 \in \N$ with $n \ge 2$. Assume: :$\displaystyle \laptrans {\map {f^{\paren {n - 1} } } t} = s^{n - 1} \laptrans {\map f t}- \sum_{j \mathop = 1}^{n - 1} s^{j - 1} \map {f^{\paren {n - 1 - j} } } 0$ This is our induction hypothesis. === Induction Step === This is our induction step: By hypothesis $f^\paren {n - 1}$ is of exponential order. Thus Laplace Transform of Derivative can be invoked: {{begin-eqn}} {{eqn | l = \laptrans {\map {f^{\paren n} } t}       | r = s \laptrans {\map {f^{\paren {n - 1} } } t} - \map f 0       | c = Laplace Transform of Derivative }} {{eqn | r = s \paren {s^{n - 1} \laptrans {\map f t} - \sum_{j \mathop = 1}^{n - 1} s^{j - 1} \map {f^{\paren {n - 1 - j} } } 0} - \map f 0       | c = Induction Hypothesis }} {{eqn | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^{n - 1} s^{j - 1 + 1} \map {f^{\paren {n - j} } } 0 - s^0 \map {f^{\paren 0} } 0       | c = {{Defof|Zeroth Derivative}} }} {{eqn | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 0}^{n - 1} s^j \map {f^{\paren {n - j} } } 0 }} {{eqn | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^n s^{j - 1} \map {f^{\paren {n - j} } } 0       | c = replace $j$ with $j - 1$ }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. {{qed}}
{{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}}
{{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}}
{{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}}
{{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}}
Let $u = a x + b$. 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 x \left({a x + b}\right)^n \ \mathrm d x       | r = \frac 1 a \int \frac {u - b} a u^n \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \frac 1 {a^2} \int u^{n+1} \ \mathrm d u - \frac b {a^2} \int u^n \ \mathrm d u       | c = Linear Combination of Integrals }} {{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 {\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       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
Let $u = a x + b$. 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 x \left({a x + b}\right)^n \ \mathrm d x       | r = \int \frac 1 a \left({\frac {u - b} a}\right)^2 u^n \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \int \frac 1 {a^3} \left({u^{n + 2} - 2 b u^{n + 1} + b^2 u^n}\right) \ \mathrm d u       | c = Square of Difference and multiplying out }} {{eqn | r = \frac 1 {a^3}  \int u^{n+3} \ \mathrm d u - \frac {2 b} {a^3} \int u^{n + 1} \ \mathrm d u + \frac {b^2} {a^3} \int u^n \ \mathrm d u       | c = Linear Combination of Integrals }} {{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 {\left({a x + b}\right)^{n + 3} } {\left({n + 3}\right) a^3} - \frac {2 b \left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^3} + \frac {b^2 \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^3} + C       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
By Distance between Element and Subset is Nonnegative: :$\map d {s, S} \ge 0$ Also, because: :$\map d {s, S} = 0$ and $s \in S$, it follows that: :$\map d {s, S} \le 0$ Hence the result. {{qed}} Category:Distance Function d5vrsa7b3u83yiv2wxrnm1vrttq59n7
By Distance between Element and Subset is Nonnegative: :$\map d {\sup S, S} \ge 0$ By definition of supremum: :$\forall \epsilon > 0: \exists s \in S: \map d {\sup S, s} < \epsilon$ meaning that, by nature of the infimum and the definition of $\map d {\sup S, S}$: :$\forall \epsilon > 0: \map d {\sup S, S} < \epsilon$ Together, these two observations lead to the conclusion that: :$\map d {\sup S, S} = 0$ as desired. {{qed}} Category:Real Analysis 9dx5tipm6gxmv12gx5qz1mcvrn3l8a6
By definition of the distance between $x$ and $S$: :$\map d {x, S} = \displaystyle \inf_{s \mathop \in S} \map d {x, s}$ From the metric space axioms: :$\forall s \in M: \map d {x, s} \ge 0$ Hence by the nature of the infimum: :$\map d {x, S} \ge 0$ as desired. {{qed}} Category:Distance Function k6h3621h7lo04cx0n3c2y9totl55ob7
By Distance between Element and Subset is Nonnegative: :$\map d {\inf S, S} \ge 0$ By definition of infimum: :$\forall \epsilon > 0: \exists s \in S: \map d {\inf S, s} < \epsilon$ meaning that, by nature of the infimum and the definition of $\map d {\inf S, S}$: :$\forall \epsilon > 0: \map d {\inf S, S} < \epsilon$ Together, these two observations lead to the conclusion that: :$\map d {\inf S, S} = 0$ as desired. {{qed}} Category:Real Analysis ge783khyr89w1rtnt8t2t2aw56bn4vm
Put $u = \sqrt{a x + b}$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {a x + b} }       | r = \frac 2 a \int u \frac {\d u} u       | c = Primitive of Function of $\sqrt {a x + b}$ }} {{eqn | r = \frac 2 a \int \rd u       | c =  }} {{eqn | r = \frac 2 a u + C       | c = Primitive of Constant }} {{eqn | r = \frac {2 \sqrt {a x + b} } a + C       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \sqrt{a x + b}       | c =  }} {{eqn | l = x       | r = \frac {u^2 - b} a       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = F \left({\sqrt{a x + b} }\right)       | r = \frac x {\sqrt{a x + b} }       | c =  }} {{eqn | ll= \implies       | l = F \left({u}\right)       | r = \left({\frac {u^2 - b} a}\right) \frac 1 u       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\sqrt{a x + b} }       | r = \frac 2 a \int u \left({\frac {u^2 - b} a}\right) \frac 1 u \ \mathrm d u       | c = Primitive of Function of $\sqrt{a x + b}$ }} {{eqn | r = \frac 2 {a^2} \int \left({u^2 - b}\right) \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 2 {a^2} \left({\frac {u^3} 3 - b u}\right) + C       | c = Primitive of Power and Primitive of Constant }} {{eqn | r = \frac 2 {a^2} \left({\frac {\left({a x + b}\right) \sqrt{a x + b} } 3 - b \sqrt{a x + b} }\right) + C       | c = substituting for $u$ }} {{eqn | r = \frac 2 {3 a^2} \left({\left({a x + b}\right) - 3 b}\right) \sqrt{a x + b} + C       | c = extracting common factors }} {{eqn | r = \frac {2 \left({a x - 2 b}\right) \sqrt{a x + b} } {3 a^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \sqrt{a x + b}       | c =  }} {{eqn | ll= \implies       | l = x       | r = \frac {u^2 - b} a       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = F \left({\sqrt{a x + b} }\right)       | r = \frac {x^2} {\sqrt{a x + b} }       | c =  }} {{eqn | ll= \implies       | l = F \left({u}\right)       | r = \left({\frac {u^2 - b} a}\right)^2 \frac 1 u       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x^2 \ \mathrm d x} {\sqrt{a x + b} }       | r = \frac 2 a \int u \left({\frac {u^2 - b} a}\right)^2 \frac 1 u \ \mathrm d u       | c = Primitive of Function of $\sqrt{a x + b}$ }} {{eqn | r = \frac 2 {a^3} \int \left({u^4 - 2 b u^2 + b^2}\right) \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 2 {a^3} \left({\frac {u^5} 5 - \frac {2 b u^3} 3 + b^2 u}\right) + C       | c = Primitive of Power and Primitive of Constant }} {{eqn | r = \frac 2 {a^3} \left({\frac {\left({a x + b}\right)^2} 5 - \frac {2 b \left({a x + b}\right)} 3 + b^2}\right)\sqrt{a x + b} + C       | c = substituting for $u$ and extracting common factors }} {{eqn | r = \frac 2 {15 a^2} \left({3 a^2 x^2 + 6 a b x + 3 b^2 - 10 a b x - 10 b^2 + 15 b^2}\right) \sqrt{a x + b} + C       | c = multiplying out and combining fractions }} {{eqn | r = \frac {2 \left({3 a^2 x^2 - 4 a b x + 8 b^2}\right) \sqrt{a x + b} } {15 a^3} + C       | c = simplifying }} {{end-eqn}} {{qed}}