hcju/proofwikips huggingface dataset:
Primitive of Reciprocal of x squared by a x + b squared/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x^2 \left({a x + b}\right)^2}$ == :$\dfrac 1 {x^2 \left({a x + b}\right)^2} \equiv -\dfrac {2 a} {b^3 x} + \dfrac 1 {b^2 x^2} + \dfrac {2 a^2} {b^3 \left({a x + b}\right)} + \dfrac {a^2} {b^2 \left({a x + b}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({a x + b}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {a x + b} + \dfrac D {\left({a x + b}\right)^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a x + b}\right)^2 + B \left({a x + b}\right)^2 + C x^2 \left({a x + b}\right) + D x^2       | c = multiplying through by $x^2 \left({a x + b}\right)^2$ }} {{eqn | o = \equiv       | r = A a^2 x^3 + 2 A a b x^2 + A b^2 x       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^2 x^2 + 2 B a b x + B b^2       | c = (tedious though this is, it helps to }} {{eqn | o =        | ro= +       | r = C a x^3 + C b x^2 + D x^2       | c = identify the equal indices) }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = D \left({-\frac b a}\right)^2       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = D       | r = \frac {a^2} {b^2}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = B b^2       | c =  }} {{eqn | n = 2       | ll= \implies       | l = B       | r = \frac 1 {b^2}       | c =  }} {{end-eqn}} Equating $1$st powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b^2 + 2 B a b       | c =  }} {{eqn | ll= \implies       | l = A b^2       | r = -\frac {2 a b} {b^2}       | c = subtituting for $B$ from $(2)$ }} {{eqn | n = 3       | ll= \implies       | l = A       | r = -\frac {2 a} {b^3}       | c =  }} {{end-eqn}} Equating $3$rd powers of $x$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + C a       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac {2 a^2} {b^3}       | c = substituting for $A$ from $(3)$ and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac {2 a} {b^3} }} {{eqn | l = B       | r = \frac 1 {b^2} }} {{eqn | l = C       | r = \frac {2 a^2} {b^3} }} {{eqn | l = D       | r = \frac {a^2} {b^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x by a x + b squared/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x \paren {a x + b}^2}$ == :$\dfrac 1 {x \paren {a x + b}^2} \equiv \dfrac 1 {b^2 x} - \dfrac a {b^2 \paren {a x + b} } - \dfrac a {b \paren {a x + b}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \paren {a x + b}^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {a x + b} + \dfrac C {\paren {a x + b}^2}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {a x + b}^2 + B x \paren {a x + b} + C x       | c = multiplying through by $x \paren {a x + b}^2$ }} {{eqn | o = \equiv       | r = A a^2 x^2 + 2 A a b x + A b^2 + B a x^2 + B b x + C x       | c = multiplying everything out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \leadsto       | l = C \paren {-\frac b a}       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \leadsto       | l = C       | r = -\frac a b       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A b^2       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = A       | r = \frac 1 {b^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + B a       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\frac a {b^2}       | c = substituting for $A$ from $(2)$ and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {b^2} }} {{eqn | l = B       | r = -\frac a {b^2} }} {{eqn | l = C       | r = -\frac a b }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x cubed by a x + b/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x^3 \left({a x + b}\right)}$ == :$\dfrac 1 {x^3 \left({a x + b}\right)} \equiv \dfrac {a^2} {b^3 x} + \dfrac {-a} {b^2 x^2} + \dfrac 1 {b x^3} + \dfrac {-a^3} {b^3 \left({a x + b}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({a x + b}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({a x + b}\right) + B x \left({a x + b}\right) + C \left({a x + b}\right) + D x^3       | c = multiplying through by $x^3 \left({a x + b}\right)$ }} {{eqn | o = \equiv       | r = A a x^3 + A b x^2 + B a x^2 + B b x + C a x + C b + D x^3       | c = multiplying everything out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = D \left({-\frac b a}\right)^3       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | n = 2       | ll= \implies       | l = D       | r = -\frac {a^3} {b^3}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = C b       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac 1 b       | c =  }} {{end-eqn}} Equating $3$rd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a + D       | c =  }} {{eqn | n = 3       | ll= \implies       | l = A       | r = \frac {a^2} {b^3}       | c = substituting for $D$ from $(2)$ and simplifying }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b + B a       | c =  }} {{eqn | r = \frac {a^2} {b^3} b + B a       | c = substituting for $A$ from $(3)$ }} {{eqn | ll= \implies       | l = B       | r = -\frac a {b^2}       | c = rearranging and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {a^2} {b^3} }} {{eqn | l = B       | r = -\frac a {b^2} }} {{eqn | l = C       | r = \frac 1 b }} {{eqn | l = D       | r = -\frac {a^3} {b^3} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x squared by a x + b/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x^2 \left({a x + b}\right)}$ == :$\dfrac 1 {x^2 \left({a x + b}\right)} \equiv -\dfrac a {b^2 x} + \dfrac 1 {b x^2} + \dfrac {a^2} {b^2 \left({a x + b}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({a x + b}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {a x + b}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a x + b}\right) + B \left({a x + b}\right) + C x^2       | c = multiplying through by $x^2 \left({a x + b}\right)$ }} {{eqn | o = \equiv       | r = A a x^2 + A b x + B a x + B b + C x^2       | c = multiplying everything out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = C \left({-\frac b a}\right)^2       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {a^2} {b^2}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = B b       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 b       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a + C       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac a {b^2}       | c = substituting for $C$ from $(2)$ and rearranging }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac a {b^2} }} {{eqn | l = B       | r = \frac 1 b }} {{eqn | l = C       | r = \frac {a^2} {b^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x by a x + b/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x \left({a x + b}\right)}$ == :$\dfrac 1 {x \left({a x + b}\right)} \equiv \dfrac 1 {b x} - \dfrac a {b \left({a x + b}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({a x + b}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {a x + b}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a x + b}\right) + B x       | c = multiplying through by $x \left({a x + b}\right)$ }} {{eqn | o = \equiv       | r = A a x + A b + B x       | c = multiplying out }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = B \left({-\frac b a}\right)       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = B       | r = -\frac a b       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A b       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 b       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 b }} {{eqn | l = B       | r = -\frac a b }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x by a x + b cubed/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x \paren {a x + b}^3}$ == :$\dfrac 1 {x \paren {a x + b}^3} \equiv \dfrac 1 {b^3 x} - \dfrac a {b^3 \paren {a x + b} } - \dfrac a {b^2 \paren {a x + b}^2} - \dfrac a {b \paren {a x + b}^3}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \paren {a x + b}^3}       | o = \equiv       | r = \dfrac A x + \dfrac B {a x + b} + \dfrac C {\paren {a x + b}^2} + \dfrac D {\paren {a x + b}^3}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {a x + b}^3 + B x \paren {a x + b}^2 + C x \paren {a x + b} + D x       | c = multiplying through by $x \paren {a x + b}^3$ }} {{eqn | o = \equiv       | r = A a^3 x^3 + 3 A a^2 b x^2 + 3 A a b^2 x + A b^3       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^2 x^3 + 2 B a b x^2 + B b^2 x + C a x^2 + C b x + D x       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \leadsto       | l = D \paren {-\frac b a}       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \leadsto       | l = D       | r = -\frac a b       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A b^3       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = A       | r = \frac 1 {b^3}       | c =  }} {{end-eqn}} Equating $3$rd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^3 + B a^2       | c =  }} {{eqn | ll= \leadsto       | l = B a^2       | r = -\frac {a^3} {b^3}       | c = substituting for $A$ from $(2)$ }} {{eqn | n = 3       | ll= \leadsto       | l = B       | r = -\frac a {b^3}       | c = simplifying }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 3 A a^2 b + 2 B a b + C a       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = -3 A a b - 2 B b       | c = rerranging }} {{eqn | r = -3 a b \frac 1 {b^3} + 2 b \frac a {b^3}       | c = substituting for $A$ from $(2)$ and $B$ from $(3)$ }} {{eqn | r = -3 \frac a {b^2} + 2 \frac a {b^2}       | c = simplifying }} {{eqn | r = -\frac a {b^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {b^3} }} {{eqn | l = B       | r = -\frac a {b^3} }} {{eqn | l = C       | r = -\frac a {b^2} }} {{eqn | l = D       | r = -\frac a b }} {{end-eqn}} Hence the result.
Primitive of Root of a x + b over Power of x/Formulation 1
Primitive of Root of a x + b over Power of x/Formulation 1 0 33874 188090 2014-05-24T06:10:32Z Prime.mover 59 Created page with "== Theorem ==  :$\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) x^{m-1} } + \frac a {2 \left({m -..." wikitext text/x-wiki == Theorem ==  :$\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) x^{m-1} } + \frac a {2 \left({m - 1}\right)} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$  == Proof == Let: {{begin-eqn}} {{eqn | l = u       | r = \sqrt{a x + b}       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u}{\mathrm d x}       | r = \frac a {2 \sqrt{a x + b} }       | c = Power Rule for Derivatives etc. }} {{eqn | l = v       | r = \frac {-1} {\left({m - 1}\right) x^{m - 1} }       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d v}{\mathrm d x}       | r = \frac 1 {x^m}       | c = Power Rule for Derivatives }} {{end-eqn}} From Integration by Parts: : $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$ from which: {{begin-eqn}} {{eqn | l = \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x       | r = \int \sqrt{a x + b} \frac 1 {x^m} \ \mathrm d x       | c =  }} {{eqn | r = \sqrt{a x + b} \frac {-1} {\left({m - 1}\right) x^{m - 1} } - \int \frac {-1} {\left({m - 1}\right) x^{m - 1} } \frac a {2 \sqrt{a x + b} } \ \mathrm d x       | c =  }} {{eqn | r = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) x^{m-1} } + \frac a {2 \left({m - 1}\right)} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }       | c = Primitive of Constant Multiple of Function }} {{end-eqn}}  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of Power of x by Root of a x + b|next=Primitive of Root of a x + b over Power of x/Formulation 2}}: $\S 14$: Integrals involving $\sqrt{a x + b}$: $14.97$
Primitive of Reciprocal of a x + b by p x + q/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of a x + b by p x + q == :$\dfrac 1 {\left({a x + b}\right) \left({p x + q}\right)} \equiv \dfrac {-a} {\left({b p - a q}\right) \left({a x + b}\right)} + \dfrac p {\left({b p - a q}\right) \left({p x + q}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\left({a x + b}\right) \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({p x + q}\right) + B \left({a x + b}\right)       | c = multiplying through by $\left({a x + b}\right) \left({p x + q}\right)$ }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = B \left({a \left({-\frac q p}\right) + b}\right)       | r = 1       | c = substituting for $x$ in $(1)$: term in $p x + q$ is $0$ }} {{eqn | ll= \implies       | l = B \left({\frac {b p - a q} p}\right)       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac p {b p - a q}       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = A \left({p \left({-\frac b a}\right) + q}\right)       | r = 1       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = A \left({\frac {-b p + a q} a}\right)       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {-a} {b p - a q}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-a} {p b - a q} }} {{eqn | l = B       | r = \frac p {b p - a q} }} {{end-eqn}} Hence the result.
Primitive of x over a x + b by p x + q/Partial Fraction Expansion
== Lemma for Primitive of x over a x + b by p x + q == :$\dfrac x {\left({a x + b}\right) \left({p x + q}\right)} \equiv \dfrac b {\left({b p - a q}\right) \left({a x + b}\right)} - \dfrac q {\left({b p - a q}\right) \left({p x + q}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac x {\left({a x + b}\right) \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = x       | o = \equiv       | r = A \left({p x + q}\right) + B \left({a x + b}\right)       | c = multiplying through by $\left({a x + b}\right) \left({p x + q}\right)$ }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = B \left({a \left({-\frac q p}\right) + b}\right)       | r = -\frac q p       | c = substituting for $x$ in $(1)$: term in $p x + q$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac {-q} {b p - a q}       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = A \left({p \left({-\frac b a}\right) + q}\right)       | r = -\frac b a       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = A       | r = \frac b {b p - a q}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac b {p b - a q} }} {{eqn | l = B       | r = \frac {-q} {b p - a q} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of a x + b squared by p x + q/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of a x + b squared by p x + q == :$\dfrac 1 {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac 1 {b p - a q} \left({\dfrac {-a p} {\left({b p - a q}\right) \left({a x + b}\right)} + \dfrac {-a} {\left({a x + b}\right)^2} + \dfrac {p^2} {\left({b p - a q}\right) \left({p x + q}\right)} }\right)$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\left({a x + b}\right)^2 \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {\left({a x + b}\right)^2} + \frac C {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a x + b}\right) \left({p x + q}\right) + B \left({p x + q}\right) + C \left({a x + b}\right)^2       | c = multiplying through by $\left({a x + b}\right)^2 \left({p x + q}\right)$ }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = B \left({p \left({-\frac b a}\right) + q}\right)       | r = 1       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac {-a} {b p - a q}       | c =  }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = C \left({a \left({-\frac q p}\right) + b}\right)^2       | r = 1       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {p^2} {\left({b p - a q}\right)^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a p + C a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = -C \frac a p       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac {p^2} {\left({b p - a q}\right)^2} \frac a p       | c = substituting for $C$ from $(2)$ }} {{eqn | ll= \implies       | l = A       | r = \frac {-a p} {\left({b p - a q}\right)^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-a p} {\left({b p - a q}\right)^2} }} {{eqn | l = B       | r = \frac {-a} {b p - a q} }} {{eqn | l = C       | r = \frac {p^2} {\left({b p - a q}\right)^2} }} {{end-eqn}} Hence the result.
Primitive of x over a x + b squared by p x + q/Partial Fraction Expansion
Primitive of x over a x + b squared by p x + q/Partial Fraction Expansion 0 33893 188130 2014-05-24T21:29:54Z Prime.mover 59 Created page with "== Lemma for Primitive of x over a x + b squared by p x + q ==  :$\dfrac x {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac 1 {b p - a q} \left({\dfrac {a q..." wikitext text/x-wiki == Lemma for Primitive of x over a x + b squared by p x + q == :$\dfrac x {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac 1 {b p - a q} \left({\dfrac {a q} {\left({b p - a q}\right) \left({a x + b}\right)} + \dfrac b {\left({a x + b}\right)^2} - \dfrac {p q} {\left({b p - a q}\right) \left({p x + q}\right)} }\right)$ == Proof == {{begin-eqn}} {{eqn | l = \frac x {\left({a x + b}\right)^2 \left({p x + q}\right)}       | o = \equiv       | r = \frac A {a x + b} + \frac B {\left({a x + b}\right)^2} + \frac C {p x + q}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = x       | o = \equiv       | r = A \left({a x + b}\right) \left({p x + q}\right) + B \left({p x + q}\right) + C \left({a x + b}\right)^2       | c = multiplying through by $\left({a x + b}\right)^2 \left({p x + q}\right)$ }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = B \left({p \left({-\frac b a}\right) + q}\right)       | r = -\frac b a       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac b {b p - a q}       | c =  }} {{end-eqn}} Setting $p x + q = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = p x + q       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac q p       | c =  }} {{eqn | ll= \implies       | l = C \left({a \left({-\frac q p}\right) + b}\right)^2       | r = -\frac q p       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {-p q} {\left({b p - a q}\right)^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a p + C a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = -C \frac a p       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac {p q} {\left({b p - a q}\right)^2} \frac a p       | c = substituting for $C$ from $(2)$ }} {{eqn | ll= \implies       | l = A       | r = \frac {a q} {\left({b p - a q}\right)^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {a q} {\left({b p - a q}\right)^2} }} {{eqn | l = B       | r = \frac b {b p - a q} }} {{eqn | l = C       | r = \frac {-p q} {\left({b p - a q}\right)^2} }} {{end-eqn}} Hence the result.