Evaluate the integral $\displaystyle \int \frac{x}{1 + x^4} dx$

If we let $u = x^2$, then

$du = 2x dx$

So,

$ \begin{equation} \begin{aligned} \int \frac{x}{1 + x^4} dx = \int \frac{x}{1 + (x^2)^2} dx &= \frac{1}{2} \int \frac{du}{1 + (u)^2}\\ \\ \text{recall that } \frac{d}{dx} \left( \tan^{-1}(x) \right) &= \frac{1}{1+x^2}\\ \\ &= \frac{1}{2} \tan^{-1} u + c\\ \\ &= \frac{1}{2} \tan^{-1} (x^2) + c \end{aligned} \end{equation} $

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