Primitive of Reciprocal of One plus Fourth Power of x

Theorem

$\ds \int \frac 1 {1 + x^4} \rd x = \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } + C$

Proof 1

From Primitive of $\dfrac {1 + x^2} {1 + x^4}$, we have:

$\ds \int \frac {x^2 + 1} {x^4 + 1} \rd x = \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C$

From Primitive of $\dfrac {-1 + x^2} {1 + x^4}$, we have:

$\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x = \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C$

We therefore have:

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

\(=\)

\(\ds \frac 1 2 \int \frac {x^2 + 1 - \paren {x^2 - 1} } {1 + x^4} \rd x\)

\(\ds \)

\(=\)

\(\ds \frac 1 2 \paren {\int \frac {x^2 + 1} {x^4 + 1} \rd x - \int \frac {x^2 - 1} {x^4 + 1} \rd x}\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } - \frac 1 2 \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } } + C\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } + C\)

Logarithm of Reciprocal

$\blacksquare$

Proof 2

A special case of Primitive of $\dfrac 1 {x^4 + a^4}$, setting $a = 1$.

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

\(=\)

\(\ds \frac 1 {4 a^3 \sqrt 2} \map \ln {\frac {x^2 + a x \sqrt 2 + a^2} {x^2 - a x \sqrt 2 + a^2} } - \frac 1 {2 a^3 \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} } + C\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \frac 1 {4 \sqrt 2} \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \frac 1 {2 \sqrt 2} \paren {\map \arctan {1 - x \sqrt 2} - \map \arctan {1 + x \sqrt 2} } + C\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \paren {\map \arctan {1 - x \sqrt 2} - \map \arctan {1 + x \sqrt 2} } } + C\)

extracting factor

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \map \arctan {\dfrac {\paren {1 - x \sqrt 2} - \paren {1 + x \sqrt 2} } {1 + \paren {1 - x \sqrt 2} \paren {1 + x \sqrt 2} } } } + C\)

Difference of Arctangents

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \map \arctan {\dfrac {-2 x \sqrt 2} {1 + \paren {1 - 2 x^2} } } } + C\)

Difference of Two Squares

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } + \map \arctan {\sqrt 2 \dfrac x {1 - x^2} } } + C\)

Inverse Tangent is Odd Function

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } + \paren {\frac \pi 2 - \map \arctan {\frac 1 {\sqrt 2} \paren {\frac {1 - x^2} x} } } } + C\)

Sum of Arctangent and Arccotangent

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \paren {\frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } + \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } + C'\)

Arctangent is Odd Function, setting $C' = \dfrac \pi 2 + C$

$\blacksquare$