Single Variable Calculus

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Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 78

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

If we let $u = x^2 $, then $\displaystyle du = 2x dx$, so $\displaystyle x dx = \frac{du}{2}$. Thus,

$ \begin{equation} \begin{aligned} \int \frac{x}{1 + x^4} dx =& \int \frac{1}{1 + x^4} \cdot xdx \\ \\ \int \frac{x}{1 + x^4} dx =& \int \frac{1}{1 + u^2} \frac{du}{2} \\ \\ \int \frac{x}{1 + x^4} dx =& \frac{1}{2} \int \frac{1}{1 + u^2} du \\ \\ \int \frac{x}{1 + x^4} dx =& \frac{1}{2} \tan^{-1} u + C \\ \\ \int \frac{x}{1 + x^4} dx =& \frac{\tan^{-1} (x^2)}{2} + C \end{aligned} \end{equation} $

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