Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 12

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Find the indefinite integral $\displaystyle \int \frac{x}{(x^2 + 1)^2} dx$

If we let $u = x^2 + 1$, then $du = 2x dx$, so $\displaystyle xdx = \frac{1}{2} du$. And

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