# The antiderivative of 1/(x^4+1): Part 4

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”: $\displaystyle \int \frac{1}{x^4 + 1} dx$

So far, I’ve shown that the denominator can be factored over the real numbers: $\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle \int \frac{dx}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)}$,

so that the technique of partial fractions can be applied. Since both quadratics in the denominator are irreducible (and the degree of the numerator is less than the degree of the denominator), the partial fractions decomposition has the form $\displaystyle \frac{1}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)} = \displaystyle \frac{Ax + B}{\left(x^2 - x \sqrt{2} + 1 \right)} + \displaystyle \frac{Cx + D}{ \left(x^2 + x \sqrt{2} + 1\right)}$

Clearing out the denominators, I get $1 = (Ax + B) \left(x^2 + x \sqrt{2} + 1\right) + (Cx + D) \left(x^2 - x \sqrt{2} + 1\right)$

or $1 = Ax^3 + Bx^2 + Ax^2 \sqrt{2} + Bx\sqrt{2} + Ax + B + Cx^3 + Dx^2 - Cx^2 \sqrt{2} - Dx\sqrt{2} + Cx + D$

or $0x^3 + 0x^2 + 0x + 1 = (A + C)x^3 + (A \sqrt{2} + B - C \sqrt{2} + D)x^2 + (A + B\sqrt{2} + C - D \sqrt{2} ) x + (B+D)$

Matching coefficients yields the following system of four equations in four unknowns: $A + C = 0$ $A\sqrt{2} + B - C\sqrt{2} + D = 0$ $A + B \sqrt{2} + C - D\sqrt{2} = 0$ $B + D = 1$

Ordinarily, four-by-four systems of linear equations are somewhat painful to solve, but this system isn’t too bad. Since $A + C = 0$ from the first equation, the third equation becomes $0 + B \sqrt{2} - D \sqrt{2} = 0$, or $B = D$.

From the fourth equation, I can conclude that $B = 1/2$ and $D = 1/2$. The second and third equations then become $A\sqrt{2} + \displaystyle \frac{1}{2} - C\sqrt{2} + \frac{1}{2} = 0$ $A + \displaystyle \frac{\sqrt{2}}{2} + C - \frac{\sqrt{2}}{2} = 0$,

or $A - C = \displaystyle -\frac{\sqrt{2}}{2}$, $A + C = 0$.

Adding the two equations yields $2A = -\displaystyle \frac{\sqrt{2}}{4}$, so that $A = -\displaystyle \frac{\sqrt{2}}{4}$ and $C = \displaystyle \frac{\sqrt{2}}{4}$.

Therefore, the integral can be rewritten as $\displaystyle \int \left( \frac{ -\displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } + \frac{ \displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 + x \sqrt{2} + 1 } \right) dx$

I’ll start evaluating this integral in tomorrow’s post.

1. #### howardat58

/  October 4, 2015

Regarding the factorisation of x^4+1 this has a lot of symmetry, and led me straight to
(x^2+ax+1)(x^2+bx+1), giving a pair of 2 variable linear equations to solve.

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