Primitive of Reciprocal of x fourth plus a fourth/Partial Fraction Expansion

Lemma for Primitive of Reciprocal of $x^4 + a^4$

$\dfrac 1 {x^4 + a^4} = \dfrac {x + a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 + a x \sqrt 2 + a^2} } - \dfrac {x - a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 - a x \sqrt 2 + a^2} }$

Proof

\(\ds \frac 1 {x^4 + a^4}\)

\(\equiv\)

\(\ds \frac 1 {\paren {x^2 + a x \sqrt 2 + a^2} \paren {x^2 - a x \sqrt 2 + a^2} }\)

Sum of Two Fourth Powers

\(\ds \)

\(\equiv\)

\(\ds \frac {A x + B} {x^2 + a x \sqrt 2 + a^2} + \frac {C x + D} {x^2 - a x \sqrt 2 + a^2}\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds \paren {A x + B} \paren {x^2 - a x \sqrt 2 + a^2} + \paren {C x + D} \paren {x^2 + a x \sqrt 2 + a^2}\)

\(\text {(1)}: \quad\)

\(\ds \)

\(\equiv\)

\(\ds A x^3 + B x^2 - A a x^2 \sqrt 2 - B a x \sqrt 2 + A x a^2 + B a^2\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds C x^3 + D x^2 + C a x^2 \sqrt 2 + D a x \sqrt 2 + C x a^2 + D a^2\)

Equating coefficients of $x^3$ in $(1)$:

\(\ds A + C\)

\(=\)

\(\ds 0\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds -A\)

\(=\)

\(\ds C\)

Equating coefficients of $x^2$ in $(1)$:

\(\ds -A a \sqrt 2 + B + C a \sqrt 2 + D\)

\(=\)

\(\ds 0\)

\(\text {(3)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds 2 C a \sqrt 2 + B + D\)

\(=\)

\(\ds 0\)

substituting for $A$ from $(2)$

Equating coefficients of $x$ in $(1)$:

\(\ds -B a \sqrt 2 + A a^2 + D a \sqrt 2 + C a^2\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds -B a \sqrt 2 - C a^2 + D a \sqrt 2 + C a^2\)

\(=\)

\(\ds 0\)

substituting for $A$ from $(2)$

\(\ds \leadsto \ \ \)

\(\ds -B + D\)

\(=\)

\(\ds 0\)

\(\text {(4)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds D\)

Setting $x = 0$ in $(1)$:

\(\ds B a^2 + D a^2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds 2 D a^2\)

\(=\)

\(\ds 1\)

substituting for $B$ from $(4)$

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

\(\ds \frac 1 {2 a^2}\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds \frac 1 {2 a^2}\)

from $(4)$

\(\ds \leadsto \ \ \)

\(\ds 2 C a \sqrt 2 + \frac 1 {a^2}\)

\(=\)

\(\ds 0\)

substituting for $B$ and $D$ in $(3)$

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds \frac {-1} {2 a^3 \sqrt 2}\)

substituting for $B$ and $D$ in $(3)$

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac 1 {2 a^3 \sqrt 2}\)

from $(2)$

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac 1 {2 a^3 \sqrt 2}\)

\(\ds B\)

\(=\)

\(\ds \frac 1 {2 a^2}\)

\(\ds C\)

\(=\)

\(\ds \frac {-1} {2 a^3 \sqrt 2}\)

\(\ds D\)

\(=\)

\(\ds \frac 1 {2 a^2}\)

Thus:

\(\ds \frac 1 {x^4 + a^4}\)

\(=\)

\(\ds \frac {\frac 1 {2 a^3 \sqrt 2} x + \frac 1 {2 a^2} } {x^2 + a x \sqrt 2 + a^2} + \frac {\frac {-1} {2 a^3 \sqrt 2} x + \frac 1 {2 a^2} } {x^2 - a x \sqrt 2 + a^2}\)

\(\ds \)

\(=\)

\(\ds \frac {x + a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 + a x \sqrt 2 + a^2} } - \frac {x - a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 - a x \sqrt 2 + a^2} }\)

$\blacksquare$