hcju/proofwikips huggingface dataset:
Primitive of x squared over a x + b squared by p x + q/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac {x^2} {\left({a x + b}\right)^2 \left({p x + q}\right)}$ == :$\dfrac {x^2} {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2 \left({a x + b}\right)} + \dfrac {-b^2} {a \left({b p - a q}\right) \left({a x + b}\right)^2} + \dfrac {q^2} {\left({b p - a q}\right)^2 \left({p x + q}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac {x^2} {\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^2       | 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^2} {a^2}       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | ll= \implies       | l = B       | r = \frac {-b^2} {a \left({b p - a q}\right)}       | 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^2} {p^2}       | c = substituting for $x$ in $(1)$: term in $a x + b$ is $0$ }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac {q^2} {\left({b p - a q}\right)^2}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = A a p + C a^2       | c =  }} {{eqn | ll= \implies       | l = A a p       | r = 1 - \frac {q^2} {\left({b p - a q}\right)^2} a^2       | c = substituting for $C$ from $(2)$ }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {a p} - \frac {a q^2} {p \left({b p - a q}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {\left({b p - a q}\right)^2 - a^2 q^2} {a p \left({b p - a q}\right)^2}       | c = common denominator }} {{eqn | ll= \implies       | l = A       | r = \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2} }} {{eqn | l = B       | r = \frac {-b^2} {a \left({b p - a q}\right)} }} {{eqn | l = C       | r = \frac {q^2} {\left({b p - a q}\right)^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x by x squared plus a squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x \paren {x^2 + a^2}$ == :$\dfrac 1 {x \paren {x^2 + a^2} } \equiv \dfrac 1 {a^2 x} - \dfrac x {a^2 \paren {x^2 + a^2} }$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \paren {x^2 + a^2} }       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {x^2 + a^2}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {x^2 + a^2} + B x^2 + C x       | c = multiplying through by $x \paren {x^2 + a^2}$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + B       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {a^2} }} {{eqn | l = B       | r = -\frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x squared by x squared plus a squared/Partial Fraction Expansion
Primitive of Reciprocal of x squared by x squared plus a squared/Partial Fraction Expansion 0 33952 188332 2014-05-30T17:22:09Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x^2 \left({x^2 + a^2}\right)$ ==  :$\dfrac 1 {x^2 \left({x^2 + a^..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \left({x^2 + a^2}\right)$ == :$\dfrac 1 {x^2 \left({x^2 + a^2}\right)} \equiv \dfrac 1 {a^2 x^2} - \dfrac 1 {a^2 \left({x^2 + a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({x^2 + a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {x^2 + a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({x^2 + a^2}\right) + B \left({x^2 + a^2}\right) + C x^3 + D x^2       | c = multiplying through by $x^2 \left({x^2 + a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = -\frac 1 {a^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x cubed by x squared plus a squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 + a^2}$ == :$\dfrac 1 {x^3 \paren {x^2 + a^2} } \equiv \dfrac 1 {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \paren {x^2 + a^2} }$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \paren {x^2 + a^2} }       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac {D x + E} {x^2 + a^2}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A x^2 \paren {x^2 + a^2} + B x \paren {x^2 + a^2} + C \paren {x^2 + a^2} + D x^4 + E x^3       | c = multiplying through by $x^3 \paren {x^2 + a^2}$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = C a^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B a^2       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + C       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = -\frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + E       | c =  }} {{eqn | ll= \leadsto       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + D       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac 1 {a^4} }} {{eqn | l = B       | r = 0 }} {{eqn | l = C       | r = \frac 1 {a^2} }} {{eqn | l = D       | r = \frac 1 {a^4} }} {{eqn | l = E       | r = 0 }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x by x squared plus a squared squared/Partial Fraction Expansion
Primitive of Reciprocal of x by x squared plus a squared squared/Partial Fraction Expansion 0 33960 188356 2014-05-31T06:41:14Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x \left({x^2 + a^2}\right)^2$ ==  :$\dfrac 1 {x \left({x^2 + a^2}..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \left({x^2 + a^2}\right)^2$ == :$\dfrac 1 {x \left({x^2 + a^2}\right)^2} \equiv \dfrac 1 {a^4 x} - \dfrac x {a^4 \left({x^2 + a^2}\right)} - \dfrac x {a^2 \left({x^2 + a^2}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({x^2 + a^2}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {x^2 + a^2} + \dfrac {D x + E} {\left({x^2 + a^2}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x^2 + a^2}\right)^2 + \left({B x + C}\right) x \left({x^2 + a^2}\right) + \left({D x + E}\right) x       | c = multiplying through by $x \left({x^2 + a^2}\right)^2$ }} {{eqn | n = 1       | o = \equiv       | r = A x^4 + 2 A a^2 x^2 + A a^4 + B x^4 + B x^2 a^2 + C x^3 + C x a^2 + D x^2 + E x       | c = multiplying everything out }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^4       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + B       | c =  }} {{eqn | ll= \implies       | l = B       | r = -\frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C + E       | c =  }} {{eqn | ll= \implies       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 A a^2 + B a^2 + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {a^4} }} {{eqn | l = B       | r = -\frac 1 {a^4} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = -\frac 1 {a^2} }} {{eqn | l = E       | r = 0 }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x squared by x squared plus a squared squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x^2 \paren {x^2 + a^2}^2$ == :$\dfrac 1 {x^2 \paren {x^2 + a^2}^2} \equiv -\dfrac 1 {a^4 x^2} - \dfrac 1 {a^4 \paren {x^2 + a^2} } - \dfrac 1 {a^2 \paren {x^2 + a^2}^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \paren {x^2 + a^2}^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {x^2 + a^2} + \dfrac {E x + F} {\paren {x^2 + a^2}^2}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A x \paren {x^2 + a^2}^2 + B \paren {x^2 + a^2}^2       | c = multiplying through by $x^2 \paren {x^2 + a^2}^2$ }} {{eqn | o =       | ro= +       | r = \paren {C x + D} x^2 \paren {x^2 + a^2} + \paren {E x + F} x^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^5 + 2 A a^2 x^3 + A a^4 x + B x^4 + 2 B a^2 x^2 + B a^4       | c = multiplying everything out }} {{eqn | o =       | ro= +       | r = C x^5 + C x^3 a^2 + D x^4 + D x^2 a^2 + E x^3 + F x^2       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^4       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + D       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = -\frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^5$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 2 A a^2 + C a^2 + E       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 B a^2 + D a^2 + F       | c =  }} {{eqn | ll= \leadsto       | l = F       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac 1 {a^4} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = -\frac 1 {a^4} }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = -\frac 1 {a^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x cubed by x squared plus a squared squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x^3 \left({x^2 + a^2}\right)^2$ == :$\dfrac 1 {x^3 \left({x^2 + a^2}\right)^2} \equiv -\dfrac 2 {a^6 x} + \dfrac 1 {a^4 x^3} + \dfrac {2 x} {a^6 \left({x^2 + a^2}\right)} + \dfrac x {a^4 \left({x^2 + a^2}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({x^2 + a^2}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac {D x + E} {x^2 + a^2} + \dfrac {F x + G} {\left({x^2 + a^2}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({x^2 + a^2}\right)^2 + B x \left({x^2 + a^2}\right)^2 + C \left({x^2 + a^2}\right)^2       | c = multiplying through by $x^2 \left({x^2 + a^2}\right)^2$ }} {{eqn | o =       | ro= +       | r = \left({D x + E}\right) x^3 \left({x^2 + a^2}\right) + \left({F x + G}\right) x^3       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^6 + 2 A a^2 x^4 + A a^4 x^2 + B x^5 + 2 B a^2 x^3 + B a^4 x       | c = multiplying everything out }} {{eqn | o =       | ro= +       | r = C x^4 + 2 C a^2 x^2 + C a^4       | c =  }} {{eqn | o =       | ro= +       | r = D x^6 + D x^4 a^2 + E x^5 + E x^3 a^2 + F x^4 + G x^3       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = C a^4       | r = 1       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^4 + 2 C a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = -\frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x^6$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 2 {a^6}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = B       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^5$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + E       | c =  }} {{eqn | ll= \implies       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 B a^2 + E a^2 + G       | c =  }} {{eqn | ll= \implies       | l = G       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 2 A a^2 + C + D a^2 + F       | r = 0       | c =  }} {{eqn | ll= \implies       | l = F       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = -\frac 2 {a^6} }} {{eqn | l = B       | r = 0 }} {{eqn | l = C       | r = \frac 1 {a^4} }} {{eqn | l = D       | r = \frac 2 {a^6} }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = \frac 1 {a^4} }} {{eqn | l = G       | r = 0 }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Proof 2/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $\left({x^2 - a^2}\right)$ == :$\dfrac 1 {x^2 - a^2} \equiv \dfrac 1 {2 a \left({x - a}\right)} - \dfrac 1 {2 a \left({x + a}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\left({x^2 - a^2}\right)}       | o = \equiv       | r = \frac A {x - a} + \frac B {x + a}       | c = Difference of Two Squares }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x + a}\right) + B \left({x - a}\right)       | c = multiplying through by $\left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 2 a + B \cdot 0       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {2 a}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 0 + B \cdot \left({-2 a}\right)       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac {-1} {2 a}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {2 a} }} {{eqn | l = B       | r = \frac {-1} {2 a} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x by x squared minus a squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x \left({x^2 - a^2}\right)$ == :$\dfrac 1 {x \left({x^2 - a^2}\right)} \equiv \dfrac {-1} {a^2 x} + \dfrac x {a^2 \left({x^2 - a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({x^2 - a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {x^2 - a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x^2 - a^2}\right) + B x^2 + C x       | c = multiplying through by $x \left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = -A a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {-1} {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + B       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {a^2} }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x squared by x squared minus a squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x^2 \left({x^2 - a^2}\right)$ == :$\dfrac 1 {x^2 \left({x^2 - a^2}\right)} \equiv \dfrac 1 {a^2 \left({x^2 - a^2}\right)} - \dfrac 1 {a^2 x^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({x^2 - a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {x^2 - a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({x^2 - a^2}\right) + B \left({x^2 - a^2}\right) + C x^3 + D x^2       | c = multiplying through by $x^2 \left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = -B a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = -\frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac {-1} {a^2} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = \frac 1 {a^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x cubed by x squared minus a squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x^3 \left({x^2 - a^2}\right)$ == :$\dfrac 1 {x^3 \left({x^2 - a^2}\right)} \equiv \dfrac {-1} {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \left({x^2 - a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({x^2 - a^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac {D x + E} {x^2 - a^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({x^2 - a^2}\right) + B x \left({x^2 - a^2}\right) + C \left({x^2 - a^2}\right) + D x^4 + E x^3       | c = multiplying through by $x^3 \left({x^2 - a^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = -C a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac {-1} {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -B a^2       | c =  }} {{eqn | ll= \implies       | l = B       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A a^2 + C       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac {-1} {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B + E       | c =  }} {{eqn | ll= \implies       | l = E       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^4$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {-1} {a^4} }} {{eqn | l = B       | r = 0 }} {{eqn | l = C       | r = \frac {-1} {a^2} }} {{eqn | l = D       | r = \frac 1 {a^4} }} {{eqn | l = E       | r = 0 }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x squared minus a squared squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $\left({x^2 - a^2}\right)^2$ == :$\dfrac 1 {\left({x^2 - a^2}\right)^2} \equiv \dfrac 1 {4 a^3 \left({x + a}\right)} - \dfrac 1 {4 a^3 \left({x - a}\right)} + \dfrac 1 {4 a^2 \left({x + a}\right)^2} + \dfrac 1 {4 a^2 \left({x - a}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({x^2 - a^2}\right)^2}       | r = \dfrac 1 {\left({x + a}\right)^2 \left({x - a}\right)^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \dfrac A {x + a} + \dfrac B {\left({x + a}\right)^2} + \dfrac C {x - a} + \dfrac D {\left({x - a}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A \left({x^2 - a^2}\right) \left({x - a}\right) + B \left({x - a}\right)^2 + C \left({x^2 - a^2}\right) \left({x + a}\right) + D \left({x + a}\right)^2       | c = multiplying through by $\left({x^2 - a^2}\right)^2$ }} {{eqn | n = 1       | o = \equiv       | r = A x^3 - A a x^2 - A a^2 x + A a^3 + B x^2 - 2 B a x + B a^2       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C x^3 + C a x^2 - C a^2 x - C a^3 + D x^2 + 2 D a x + D a^2       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \left({-2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A + C       | c =  }} {{eqn | ll= \implies       | l = A       | r = -C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = - A a + C a + B + D       | r = 0       | c =  }} {{eqn | ll= \implies       | l = - A a + C a + \frac 1 {4 a^2} + \frac 1 {4 a^2}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = C a + C a + \frac 1 {2 a^2}       | r = 0       | c = as $A = - C$ }} {{eqn | ll= \implies       | l = 2 C        | r = \frac {-1} {2 a^3}       | c =  }} {{eqn | ll= \implies       | l = C        | r = \frac {-1} {4 a^3}       | c =  }} {{eqn | ll= \implies       | l = A        | r = \frac 1 {4 a^3}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {4 a^3} }} {{eqn | l = B       | r = \frac 1 {4 a^2} }} {{eqn | l = C       | r = \frac {-1} {4 a^3} }} {{eqn | l = D       | r = \frac 1 {4 a^2} }} {{end-eqn}} Hence the result.