hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | o =        | r = \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} }       | c =  }} {{eqn | r = \int \paren {\frac 1 {b p - a q} \paren {\frac {a q} {\paren {b p - a q} \paren {a x + b} } + \frac b {\paren {a x + b}^2} - \frac {p q} {\paren {b p - a q} \paren {p x + q} } } } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac {a q} {b p - a q} \int \frac {\d x} {a x + b} + b \int \frac {\d x} {\paren {a x + b}^2} - \frac {p q} {b p - a q} \int \frac {\d x} {p x + q} }       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac {a q} {b p - a q} \frac 1 a \ln \size {a x + b} + b \int \frac {\d x} {\paren {a x + b}^2} + \frac {p q} {b p - a q} \frac 1 p \ln \size {p x + q} } + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {a x + b} + b \frac {-1} {a \paren {a x + b} } - \frac q {b p - a q} \ln \size {p x + q} } + C       | c = Primitive of $\dfrac 1 {\left({a x + b}\right)^2}$ }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)}       | c =  }} {{eqn | r = \int \left({\frac {-b^2} {a \left({b p - a q}\right) \left({a x + b}\right)^2} + \frac {q^2} {\left({b p - a q}\right)^2 \left({p x + q}\right)} + \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2 \left({a x + b}\right)} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac {-b^2} {a \left({b p - a q}\right)} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {q^2} {\left({b p - a q}\right)^2} \int \frac {\mathrm d x} {p x + q} + \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2} \int \frac {\mathrm d x} {a x + b}       | c = Linear Combination of Integrals }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^2}       | r = \frac {-1} {a \left({a x + b}\right)}       | c = Primitive of Reciprocal of $\left({a x + b}\right)^2$ }} {{eqn | l = \int \frac {\mathrm d x} {p x + q}       | r = \frac 1 p \ln \left \vert {p x + q} \right \vert       | c = Primitive of Reciprocal of $a x + b$ }} {{eqn | l = \int \frac {\mathrm d x} {a x + b}       | r = \frac 1 a \ln \left \vert {a x + b} \right \vert       | c = Primitive of Reciprocal of $a x + b$ }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | o =        | r = \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)}       | c =  }} {{eqn | r = \frac {-b^2} {a \left({b p - a q}\right)} \frac {-1} {a \left({a x + b}\right)} + \frac {q^2} {\left({b p - a q}\right)^2} \left({\frac 1 p \ln \left\vert{p x + q}\right\vert}\right) + \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2} \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) + C       | c = Substitution }} {{eqn | r = \frac {b^2} {\left({b p - a q}\right) a^2 \left({a x + b}\right)} + \frac 1 {\left({b p - a q}\right)^2} \left({\frac {q^2} p \ln \left\vert{p x + q}\right\vert + \frac {b \left({b p - 2 a q}\right)} {a^2} \ln \left\vert{a x + b}\right\vert}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
=== Smooth functions with compact support === {{explain|In several places in the below, this construct was seen: $\vert \nabla u (x_1, \dots, x_{m - 1}, t\vert^2$ where it appears that the close of the round bracket parenthesis "$)$" was been omitted. It has been assmued in the corrected version that it should be this: $\vert\nabla u (x_1,\dots,x_{m-1},t)\vert^2$ but it is not obvious and may require someone knowledgeable to check this out.}} Let $u \in \map {C_0^\infty} G$. Put $\map u x = 0$ if $x = \tuple {x_1, x_2, \ldots, x_n} \notin G$. Then: :$u \in \map {C_0^\infty} {\R^n}$ Denote by $a,b$ the extremes of the last coordinate: $a = \inf \limits_{x \mathop \in G} x_n$ and $b = \sup \limits_{y \mathop \in G} x_n$ Then for any $x$: :$\displaystyle \size {\map u {x_1, x_2, \ldots, x_n} }^2 = \size {\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \tuple {x_1, x_2, \ldots, x_{n - 1}, t} \rd t}^2$ By Cauchy-Bunyakovsky-Schwarz Inequality: {{begin-eqn}} {{eqn | l = \left\vert{\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \left({x_1, \ldots, x_{n-1}, t}\right) \rd t}\right\vert^2       | o = \le       | r = \int \limits_a^{x_n} 1^2 \ \mathrm d t \int \limits_a^{x_n} \left\vert{\frac{\partial u} {\partial x_n} \left({x_1, \ldots, x_{n-1}, t}\right)}\right\vert^2 \rd t       | c =  }} {{eqn | o = \le       | r = \map {\operatorname {diam} } G \int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t       | c =  }} {{end-eqn}} Integrating this we get: :$\displaystyle \norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x$ By Fubini's Theorem: {{begin-eqn}} {{eqn | l = \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x       | r = \int \limits_a^b \rd x_m \int \limits_{\mathbb R^n} \size {\map {\nabla u} {x_1, \ldots, x_{n - 1}, x_m} }^2 \rd x       | c =  }} {{eqn | o = \le       | r = \map {\operatorname {diam}^2} G \norm {\nabla u}^2_{\map {L^2} G}       | c =  }} {{end-eqn}} {{qed}} === General case === Let now $u \in \map {W^{2, 1}_0} G$. There exists a sequence $\displaystyle \sequence {u_n}_{n \mathop = 1}^\infty \subset \map {C_0^\infty} G$ such that: :$\norm {u - u_n}_{\map {W^{2, 1} } G} \to 0$ as $n \to \infty$. Then: :$\norm {u - u_n}_{\map {L^2} G} \to 0$ and: :$\norm {\nabla u - \nabla u_n}_{\map {L^2} G} \to 0$ As $\size {\, \norm {u - u_n} \,} \le \norm {u - u_n}$, it follows that: :$\norm {u_n}_{\map {L^2} G} \to \norm u_{\map {L^2} G}$ and: :$\norm {\nabla u_n}_{\map {L^2} G} \to \norm {\nabla u}_{\map {L^2} G}$ Since the inequality is correct for all $u_n$, it is also correct for $u$. {{qed}} {{Namedfor|Kurt Otto Friedrichs|cat = Friedrichs}} 13j07onk2aoc8ytc21fm6g3towbg039
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power: :$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)$ Setting $m := -m$ and $n := -n$: {{begin-eqn}} {{eqn | o =        | r = \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}       | c =  }} {{eqn | r = \int \left({a x + b}\right)^{-m} \left({p x + q}\right)^{-n} \ \mathrm d x       | c =  }} {{eqn | r = \frac 1 {\left({-n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{-m+1} \left({p x + q}\right)^{-n+1} - a \left({-m - n + 2}\right) \int \left({a x + b}\right)^{-m} \left({p x + q}\right)^{-n+1} \ \mathrm d x}\right)       | c =  }} {{eqn | r = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)       | c =  }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power: :$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$ Putting $n + 1$ for $n$, this yields: {{begin-eqn}} {{eqn | l = \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x       | r = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} } {\left({m + n + 2}\right) a} - \frac {\left({n + 1}\right) \left({b p - a q}\right)} {\left({m + n + 2}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x       | c =  }} {{eqn | ll= \implies       | l = \int \left({a x + b}\right)^m \left({p x + q}\right)^n       | r = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} } {\left({n + 1}\right) \left({b p - a q}\right)} - \frac {\left({m + n + 2}\right) a} {\left({n + 1}\right) \left({b p - a q}\right)} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x       | c =  }} {{eqn | r = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {a x + b} {p x + q} \ \mathrm d x       | r = a \int \frac {x \ \mathrm d x} {p x + q} + b \int \frac {\mathrm d x} {p x + q}       | c = Linear Combination of Integrals }} {{eqn | r = a \left({\frac x p - \frac q {p^2} \ln \left\vert{p x + q}\right\vert}\right) + b \int \frac {\mathrm d x} {p x + q} + C       | c = Primitive of $x$ over $a x + b$ }} {{eqn | r = \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + b \left({\frac 1 p \ln \left\vert{p x + q}\right\vert}\right) + C       | c = Primitive of Reciprocal of $a x + b$ }} {{eqn | r = \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + \frac b p \ln \left\vert{p x + q}\right\vert + C       | c =  }} {{eqn | r = \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C       | c = common denominator }} {{end-eqn}} {{qed}}
Let $s \in \Z$.  {{begin-eqn}} {{eqn | l = v       | r = x^s       | c =  }} {{eqn | n = 1       | ll= \implies       | l = \frac {\d v} {\d x}       | r = s x^{s - 1}       | c = Power Rule for Derivatives }} {{end-eqn}} Let $u \dfrac {\d v} {\d x} = x^m \left({a x + b}\right)^n$. Then: {{begin-eqn}} {{eqn | l = u       | r = \frac {x^m \left({a x + b}\right)^n} {s x^{s - 1} }       | c = from $(1)$ }} {{eqn | r = \frac {x^{m - s + 1} \left({a x + b}\right)^n} s       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac {x^{m - s + 1} a n \left({a x + b}\right)^{n - 1} } s + \frac {\left({m - s + 1}\right) x^{m - s} \left({a x + b}\right)^n} s       | c = Primitive of Power of $a x + b$ and Product Rule for Derivatives }} {{eqn | r = \frac {x^{m - s} } s \left({a x + b}\right)^{n - 1} \left({a n x + \left({m - s + 1}\right) \left({a x + b}\right) }\right)       | c = extracting $\dfrac {x^{m - s} \left({a x + b}\right)^n} s$ as a factor }} {{eqn | r = \frac {x^{m - s} } s \left({a x + b}\right)^{n - 1} \left({a n x + a m x - a s x + a x + m b - s b + b}\right)       | c = multiplying out }} {{eqn | n = 2       | r = \frac {x^{m - s} } s \left({a x + b}\right)^{n - 1} \left({\left({m + n + 1 - s}\right) a x + \left({m - s + 1}\right) b}\right)       | c = gathering terms }} {{end-eqn}} Let $s$ be selected such that $m + n + 1 - s = 0$. Then $s = m + n + 1$. Thus $(2)$ after rearrangement becomes: :$\dfrac {\d u} {\d x} = \dfrac {x^{m - s} } {m + n + 1} \left({a x + b}\right)^{n - 1} \left({\left({m - s + 1}\right) b}\right)$ Then: {{begin-eqn}} {{eqn | l = u v       | r = \frac {x^{m - s + 1} \left({a x + b}\right)^n} {m + n + 1} x^s       | c =  }} {{eqn | r = \frac {x^{m + 1} \left({a x + b}\right)^n} {m + n + 1}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = v \frac {\d u} {\d x}       | r = x^s \dfrac {x^{m - s} } {m + n + 1} \left({a x + b}\right)^{n - 1} \left({\left({m - s + 1}\right) b}\right)       | c =  }} {{eqn | r = \dfrac {x^m} {m + n + 1} \left({a x + b}\right)^{n - 1} \left({\left({m - \left({m + n + 1}\right) + 1}\right) b}\right)       | c =  }} {{eqn | r = \dfrac {x^m} {m + n + 1} \left({a x + b}\right)^{n - 1} \left({-n b}\right)       | c =  }} {{end-eqn}} Thus by Integration by Parts: :$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \rd x$ {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power: :$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$ Setting $a := 1, b := 0, p x + q := a x + b$: {{begin-eqn}} {{eqn | l = \int x^m \paren {a x + b}^n \rd x       | r = \frac {\paren {1 x + 0}^{m + 1} \paren {a x + b}^n} {\paren {m + n + 1} 1} - \frac {n \paren {0 a - 1 b} } {\paren {m + n + 1} 1} \int \paren {1 x + 0}^m \paren {a x + b}^{n - 1} \rd x       | c =  }} {{eqn | r = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} - \frac {n b} {\paren {m + n + 1} } \int x^m \paren {a x + b}^{n - 1} \rd x       | c =  }} {{end-eqn}} {{qed}}
Let $s \in \Z$.  {{begin-eqn}} {{eqn | l = v       | r = \left({a x + b}\right)^s       | c =  }} {{eqn | n = 1       | ll= \implies       | l = \frac {\d v} {\d x}       | r = a s \left({a x + b}\right)^{s - 1}       | c = Power Rule for Derivatives and Derivatives of Function of $a x + b$ }} {{end-eqn}} Let $u \dfrac {\d v} {\d x} = x^m \left({a x + b}\right)^n$. Then: {{begin-eqn}} {{eqn | l = u       | r = \frac {x^m \left({a x + b}\right)^n} {a s \left({a x + b}\right)^{s - 1} }       | c = from $(1)$ }} {{eqn | r = \frac {x^m} {a s} \left({a x + b}\right)^{n - s + 1}       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac {m x^{m - 1} } {a s} \left({a x + b}\right)^{n - s + 1} + \frac {x^m} {a s} a \left({n - s + 1}\right) \left({a x + b}\right)^{n - s}       | c = above rules and Product Rule for Derivatives }} {{eqn | r = \frac {x^{m - 1} } {a s} \left({a x + b}\right)^{n - s} \left({m \left({a x + b}\right) + a \left({n - s + 1}\right) x}\right)       | c = extracting $\dfrac {x^{m - 1} } {a s} \left({a x + b}\right)^{n - s}$ as a factor }} {{eqn | n = 2       | r = \frac {x^{m - 1} } {a s} \left({a x + b}\right)^{n - s} \left({m b + a \left({m + n - s + 1}\right) x}\right)       | c = rearranging }} {{end-eqn}} Let $s$ be selected such that $m + n + 1 - s = 0$. Then $s = m + n + 1$. Thus $(2)$ after rearrangement becomes: :$\dfrac {\d u} {\d x} = \dfrac {m b x^{m - 1} \left({a x + b}\right)^{n - s} } {\left({m + n + 1}\right) a}$ Then: {{begin-eqn}} {{eqn | l = u v       | r = \frac {x^m} {a s} \left({a x + b}\right)^{n - s + 1} \left({a x + b}\right)^s       | c =  }} {{eqn | r = \frac {x^m \left({a x + b}\right)^{n + 1} } {\left({m + n + 1}\right) a}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = v \frac {\d u} {\d x}       | r = \left({a x + b}\right)^s \frac {m b x^{m - 1} \left({a x + b}\right)^{n - s} } {\left({m + n + 1}\right) a}       | c =  }} {{eqn | r = \frac {m b x^{m - 1} \left({a x + b}\right)^n} {\left({m + n + 1}\right) a}       | c =  }} {{end-eqn}} Thus by Integration by Parts: :$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^m \left({a x + b}\right)^{n + 1} } {\left({m + n + 1}\right) a} - \frac {m b} {\left({m + n + 1}\right) a} \int x^{m - 1} \left({a x + b}\right)^n \rd x$ {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power: :$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$ Setting $p := 1, q := 0, n := m, m := n$: {{begin-eqn}} {{eqn | l = \int x^m \paren {a x + b}^n \rd x       | r = \frac {\paren {a x + b}^{n + 1} \paren {1 x + 0}^m} {\paren {m + n + 1} a} - \frac {m \paren {b 1 - a 0} } {\paren {m + n + 1} a} \int \paren {a x + b}^n \paren {1 x + 0}^{m - 1} \rd x       | c =  }} {{eqn | r = \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       | c =  }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$: :$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$ Substituting $n + 1$ for $n$: {{begin-eqn}} {{eqn | l = \int x^m \paren {a x + b}^{n + 1} \rd x       | r = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \frac {\paren {n + 1} b} {m + n + 2} \int x^m \paren {a x + b}^n \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\paren {n + 1} b} {m + n + 2} \int x^m \paren {a x + b}^n \rd x       | r = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \int x^m \paren {a x + b}^{n + 1} \rd x       | c = rearrangement }} {{eqn | ll= \leadsto       | l = \int x^m \paren {a x + b}^n \rd x       | r = \frac {m + n + 2} {\paren {n + 1} b} \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {m + n + 2} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x       | c = rearrangement }} {{eqn | ll= \leadsto       | l = \int x^m \paren {a x + b}^n \rd x       | r = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x       | c = rearrangement }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power: :$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - a \paren {m + n + 2} \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$ Setting $a := 1, b := 0, p x + q := a x + b$: {{begin-eqn}} {{eqn | l = \int x^m \paren {a x + b}^n \rd x       | r = \frac 1 {\paren {n + 1} \paren {0 a - 1 b} } \paren {\paren {1 x + 0}^{m + 1} \paren {a x + b}^{n + 1} - 1 \paren {m + n + 2} \int \paren {1 x + 0}^m \paren {a x + b} ^{n + 1} \rd x}       | c =  }} {{eqn | r = \frac 1 {\paren {n + 1} \paren {-b} } \paren {x^{m + 1} \paren {a x + b}^{n + 1} - \paren {m + n + 2} \int x^m \paren {a x + b}^{n + 1} \rd x}       | c =  }} {{eqn | r = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x       | c =  }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$: :$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^m \left({a x + b}\right)^{n + 1} } {\left({m + n + 1}\right) a} - \frac {m b} {\left({m + n + 1}\right) a} \int x^{m - 1} \left({a x + b}\right)^n \rd x$ Substituting $m + 1$ for $m$: {{begin-eqn}} {{eqn | l = \int x^{m + 1} \left({a x + b}\right)^n \rd x       | r = \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + n + 2}\right) a} - \frac {\left({m + 1}\right) b} {\left({m + n + 2}\right) a} \int x^m \left({a x + b}\right)^n \rd x       | c =  }} {{eqn | ll= \implies       | l = \frac {\left({m + 1}\right) b} {\left({m + n + 2}\right) a} \int x^m \left({a x + b}\right)^n \rd x       | r = \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + n + 2}\right) a} - \int x^{m + 1} \left({a x + b}\right)^n \rd x       | c = rearranging }} {{eqn | ll= \implies       | l = \int x^m \left({a x + b}\right)^n \rd x       | r = \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + n + 2}\right) a} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x       | c = rearranging }} {{eqn | r = \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x       | c = rearranging }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power: :$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \rd x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m + 1} \left({p x + q}\right)^{n + 1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n + 1} \rd x}\right)$ Setting $p := 1, q := 0, m := n, n := m$: {{begin-eqn}} {{eqn | l = \int x^m \left({a x + b}\right)^n \rd x       | r = \frac 1 {\left({m + 1}\right) \left({b 1 - a 0}\right)} \left({\left({a x + b}\right)^{n + 1} \left({1 x + 0}\right)^{m + 1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^n \left({0 x + 1}\right)^{m + 1} \rd x}\right)       | c =  }} {{eqn | r = \frac 1 {\left({m + 1}\right) b} \left({\left({a x + b}\right)^{n + 1} x^{m+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^n x^{m + 1} \rd x}\right)       | c =  }} {{eqn | r = \frac {x^{m + 1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x       | c =  }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power: :$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \rd x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m + 1} \left({p x + q}\right)^{n + 1} - \left({m + n + 2}\right) a \int \left({a x + b}\right)^m \left({p x + q}\right)^{n + 1} \rd x}\right)$ Setting $n := -n$: {{begin-eqn}} {{eqn | o =        | r = \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n} \rd x       | c =  }} {{eqn | r = \int \left({a x + b}\right)^m \left({p x + q}\right)^{-n} \rd x       | c =  }} {{eqn | r = \frac 1 {\left({-n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m + 1} \left({p x + q}\right)^{-n + 1} - \left({m + -n + 2}\right) a \int \left({a x + b}\right)^m \left({p x + q}\right)^{-n + 1} \rd x}\right)       | c =  }} {{eqn | r = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac {\left({a x + b}\right)^{m + 1} } {\left({p x + q}\right)^{n - 1} } + \left({n - m - 2}\right) a \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^{n - 1} } }\right)       | c =  }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power: :$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} a} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$ Setting $n := -n$: {{begin-eqn}} {{eqn | o =        | r = \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x       | c =  }} {{eqn | r = \int \paren {a x + b}^m \paren {p x + q}^{-n} \rd x       | c =  }} {{eqn | r = \frac {\paren {a x + b}^m \paren {p x + q}^{-n + 1} } {\paren {m - n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m - n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^{-n} \rd x       | c =  }} {{eqn | r = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m  \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}       | c =  }} {{end-eqn}} {{qed}}