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

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
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\int \frac{x}{(x^2 + 1)^2} dx =& \int \frac{1}{u^2} \frac{1}{2} du
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\int \frac{x}{(x^2 + 1)^2} dx =& \frac{1}{2} \int \frac{1}{u^2} du
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\int \frac{x}{(x^2 + 1)^2} dx =& \frac{1}{2} \int u^{-2} du
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\int \frac{x}{(x^2 + 1)^2} dx =& \frac{1}{2} \cdot \frac{u^{-2 + 1}}{-2 + 1} + C
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\int \frac{x}{(x^2 + 1)^2} dx =& \frac{1}{2} \cdot \frac{u^{-1}}{-1} + C
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\int \frac{x}{(x^2 + 1)^2} dx =& \frac{-1}{2u} + C
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\int \frac{x}{(x^2 + 1)^2} dx =& \frac{-1}{2 (x^2 + 1)} + C


\end{aligned}
\end{equation}
$