An indefinite integral

Evaluate the integral

$\mathcal{J} = \int \frac{\mathrm{d}x}{1+x^4}$

Solution

We note that:

$\begin{align*} \frac{1}{x^4+1} &= \frac{1}{2} \cdot \left ( \frac{1+x^{-2}}{x^2+x^{-2}} - \frac{1-x^{-2}}{x^2+x^{-2}} \right ) \\ &=\frac{1}{2} \cdot \left ( \frac{1+x^{-2}}{\left ( x-x^{-1} \right )^2+2} - \frac{1-x^{-2}}{\left ( x+x^{-1} \right )^2-2} \right ) \end{align*}$

If we set $x-x^{-1}=t$ and $x+x^{-1}=s$ then:

$\begin{align*} \mathcal{J} &= \int \frac{\mathrm{d}x}{x^4+1} \\ &=\frac{1}{2}\left ( \int \frac{\mathrm{d}t}{t^2+2} - \int \frac{\mathrm{d}s}{s^2-2} \right ) \\ &= \frac{1}{2\sqrt{2}} \arctan \frac{t}{\sqrt{2}} - \frac{1}{4\sqrt{2}} \left | \frac{s-\sqrt{2}}{s+\sqrt{2}} \right | \\ &= \frac{1}{2\sqrt{2}} \arctan \frac{x^2-1}{\sqrt{2}x} - \frac{1}{4\sqrt{2}} \ln \left ( \frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1} \right ) + c \; , \; c \in \mathbb{R} \end{align*}$

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An indefinite integral

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