Primitive of Reciprocal of x squared by x squared minus a squared squared/Partial Fraction Expansion
Primitive of Reciprocal of x squared by x squared minus a squared squared/Partial Fraction Expansion 0 34005 188492 2014-06-03T15:13:21Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x^2 \left({x^2 - a^2}\right)^2$ ==  :$\dfrac 1 {x^2 \lef..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \left({x^2 - a^2}\right)^2$ == :$\dfrac 1 {x^2 \left({x^2 - a^2}\right)^2} \equiv \dfrac 1 {a^4 x^2} + \dfrac 3 {4 a^5 \left({x + a}\right)} - \dfrac 3 {4 a^5 \left({x - a}\right)} + \dfrac 1 {4 a^4 \left({x + a}\right)^2} + \dfrac 1 {4 a^4 \left({x - a}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^2 \left({x^2 - a^2}\right)^2}       | r = \frac 1 {x^2 \left({x + a}\right)^2 \left({x - a}\right)^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \frac A {x + a} + \frac B {\left({x + a}\right)^2} + \frac C {x - a} + \frac D {\left({x - a}\right)^2} + \frac E x + \frac F {x^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({x^2 - a^2}\right) \left({x - a}\right) + B x^2 \left({x - a}\right)^2       | c = multiplying through by $x^2 \left({x^2 - a^2}\right)^2$ }} {{eqn | o =        | ro= +       | r = C x^2 \left({x^2 - a^2}\right) \left({x + a}\right) + D x^2 \left({x + a}\right)^2       | c =  }} {{eqn | o =        | ro= +       | r = E x \left({x^2 - a^2}\right)^2 + F \left({x^2 - a^2}\right)^2       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^5 - A a x^4 - A a^2 x^3 + A a^3 x^2 + B x^4 - 2 B a x^3 + B a^2 x^2       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C x^5 + C a x^4 - C a^2 x^3 - C a^3 x^2 + D x^4 + 2 D a x^3 + D a^2 x^2       | c =  }} {{eqn | o =        | ro= +       | r = E x^5 - 2 E a^2 x^3 + E a^4 x + F x^4 - 2 F a^2 x^2 + F a^4       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D a^2 \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {4 a^4}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \left({-a}\right)^2 \left({-2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {4 a^4}       | c =  }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = F a^4       | r = 1       | c =  }} {{eqn | ll= \implies       | l = F       | r = \frac 1 {a^4}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = E       | c =  }} {{end-eqn}} Equating coefficients of $x^5$ in $(1)$: {{begin-eqn}} {{eqn | n = 2       | l = 0       | r = A + C + E       | c =  }} {{eqn | ll= \implies       | l = 0       | r = A + C       | c = as $E = 0$ }} {{eqn | ll= \implies       | l = A       | r = - C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^3 + B a^2 - C a^3 + D a^2 - 2 F a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = A a^3 + \frac 1 {4 a^4} a^2 - C a^3 + \frac 1 {4 a^4} a^2 - 2 \frac 1 {a^4} a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = A + \frac 1 {4 a^5} - C + \frac 1 {4 a^5} - \frac 8 {4 a^5}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = A - C       | r = \frac {8 - 1 - 1} {4 a^5}       | c =  }} {{eqn | r = \frac 3 {2 a^5}       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 3 {4 a^5}       | c =  }} {{eqn | ll= \implies       | l = C       | r = \frac {-3} {4 a^5}       | c = as $A = - C$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 3 {4 a^5} }} {{eqn | l = B       | r = \frac 1 {4 a^4} }} {{eqn | l = C       | r = \frac {-3} {4 a^5} }} {{eqn | l = D       | r = \frac 1 {4 a^4} }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = \frac 1 {a^4} }} {{end-eqn}} Hence the result.