Primitive of Reciprocal of x fourth plus a fourth/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x^4 + a^4$ == :$\dfrac 1 {x^4 + a^4} = \dfrac {x + a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 + a x \sqrt 2 + a^2}\right)} - \dfrac {x - a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 - a x \sqrt 2 + a^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^4 + a^4}       | o = \equiv       | r = \frac 1 {\left({x^2 + a x \sqrt 2 + a^2}\right) \left({x^2 - a x \sqrt 2 + a^2}\right)}       | c = Sum of Two Fourth Powers }} {{eqn | o = \equiv       | r = \frac {A x + B} {x^2 + a x \sqrt 2 + a^2} + \frac {C x + D} {x^2 - a x \sqrt 2 + a^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = \left({A x + B}\right) \left({x^2 - a x \sqrt 2 + a^2}\right) + \left({C x + D}\right) \left({x^2 + a x \sqrt 2 + a^2}\right)       | c =  }} {{eqn | n = 1       | o = \equiv       | r = A x^3 + B x^2 - A a x^2 \sqrt 2 - B a x \sqrt 2 + A x a^2 + B a^2       | c =  }} {{eqn | o =        | ro= +       | r = C x^3 + D x^2 + C a x^2 \sqrt 2 + D a x \sqrt 2 + C x a^2 + D a^2       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = A + C       | r = 0       | c =  }} {{eqn | n = 2       | ll= \implies       | l = -A       | r = C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = -A a \sqrt 2 + B + C a \sqrt 2 + D       | r = 0       | c =  }} {{eqn | n = 3       | ll= \implies       | l = 2 C a \sqrt 2 + B + D       | r = 0       | c = substituting for $A$ from $(2)$ }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = -B a \sqrt 2 + A a^2 + D a \sqrt 2 + C a^2       | r = 0       | c =  }} {{eqn | ll= \implies       | l = -B a \sqrt 2 - C a^2 + D a \sqrt 2 + C a^2       | r = 0       | c = substituting for $A$ from $(2)$ }} {{eqn | ll= \implies       | l = -B + D       | r = 0 }} {{eqn | n = 4       | ll= \implies       | l = B       | r = D }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^2 + D a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = 2 D a^2       | r = 1       | c = substituting for $B$ from $(4)$ }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {2 a^2}       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {2 a^2}       | c = from $(4)$ }} {{eqn | ll= \implies       | l = 2 C a \sqrt 2 + \frac 1 {a^2}       | r = 0       | c = substituting for $B$ and $D$ in $(3)$ }} {{eqn | ll= \implies       | l = C       | r = \frac {-1} {2 a^3 \sqrt 2}       | c = substituting for $B$ and $D$ in $(3)$ }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {2 a^3 \sqrt 2}       | c = from $(2)$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {2 a^3 \sqrt 2} }} {{eqn | l = B       | r = \frac 1 {2 a^2} }} {{eqn | l = C       | r = \frac {-1} {2 a^3 \sqrt 2} }} {{eqn | l = D       | r = \frac 1 {2 a^2} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 {x^4 + a^4}       | r = \frac {\frac 1 {2 a^3 \sqrt 2} x + \frac 1 {2 a^2} } {x^2 + a x \sqrt 2 + a^2} + \frac {\frac {-1} {2 a^3 \sqrt 2} x + \frac 1 {2 a^2} } {x^2 - a x \sqrt 2 + a^2} }} {{eqn | r = \frac {x + a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 + a x \sqrt 2 + a^2}\right)} - \frac {x - a \sqrt 2} {2 a^3 \sqrt 2 \left({x^2 - a x \sqrt 2 + a^2}\right)} }} {{end-eqn}}