求 $\int_0^2\frac{x}{(x^2-2x+2)^2}\,\mathrm dx$

isee

源自知乎提问区

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求 $\displaystyle\int_0^2\frac{x}{(x^2-2x+2)^2}\,\mathrm dx$ .

此处定积分可直接三角换元. \begin{align*}
\int_0^2\frac{x}{(x^2-2x+2)^2}\,\mathrm dx&=\int_0^2\frac{x}{\big(1+(x-1)^2\big)^2}\,\mathrm dx,
\end{align*} 令 $x-1=\tan u$ ，则 \begin{align*}
\int_0^2\frac{x}{(x^2-2x+2)^2}\,\mathrm dx&=\int_{-\frac\pi4}^{\frac\pi4}\frac{1+\tan u}{\big(1+\tan^2 u\big)^2}\sec^2 u\,\mathrm du\\[1em]
&=\int_{-\frac\pi4}^{\frac\pi4}\frac{1+\tan u}{\sec^2 u}\,\mathrm du\\[1em]
&=\int_{-\frac\pi4}^{\frac\pi4}\left(\frac12\sin 2u+\cos^2u\right)\,\mathrm du\\[1em]
{\color{blue}{(\text{奇函数}\,+\,\text{偶函数})}}\quad&=0+2\int_{0}^{\frac\pi4}\cos^2u\,\mathrm du\\[1em]
&=\int_{0}^{\frac\pi4}(1+\cos 2u)\,\mathrm du\\[1em]
&=\frac{\pi}4+\frac12.
\end{align*}

abababa

其实首先变成对称区间就好办了，令$t=x-1$，
\[原式=\int_{-1}^{1}(\frac{t}{(t^2+1)^2}+\frac{1}{(t^2+1)^2})dt\]
而$\int_{-1}^{1}\frac{t}{(t^2+1)^2}dt$是奇函数在对称区间上的积分，就是零，所以
\[原式=\int_{-1}^{1}\frac{1}{(t^2+1)^2}dt=[\frac{1}{2}(\frac{t}{t^2+1}+\arctan t)]_{-1}^{1}\]