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源自知乎提问
题： $\displaystyle \int\frac{\sqrt{x^2+2}}{x^2+1}\,\mathrm dx$ .
先分子有理化拆出熟悉部分：
\begin{align*}
\int\frac{\sqrt{x^2+2}}{x^2+1}\,\mathrm dx&=\int\frac{(x^2+1)+1}{(x^2+1)\sqrt{x^2+2}}\,\mathrm dx\\[1ex]
&=\int\frac1{\sqrt{x^2+2}}\,\mathrm dx+\int\frac1{(x^2+1)\sqrt{x^2+2}}\,\mathrm dx\\[1ex]
&=\ln\big(x+\sqrt{x^2+2}\big)+\color{red}{\int\frac1{(x^2+1)\sqrt{x^2+2}}\,\mathrm dx}
\end{align*}对红色部分尝试欧拉代换：
令 $\color{blue}{\sqrt {x^2+2}=t-x}$，亦即 $t=x+\sqrt{x^2+2}$ ，则 $x=\frac{t^2-2}{2t}$ ，于是 \[\sqrt{x^2+2}=t-\frac{t^2-2}{2t}=\frac{t^2+2}{2t},\]以及 \[\mathrm dx=\frac{4t^2-(t^2-2)\cdot 2}{4t^2}\,\mathrm dt=\frac{t^2+2}{2t^2}\,\mathrm dt.\] 从而
\begin{align*}
\color{red}{\int\frac1{(x^2+1)\sqrt{x^2+2}}\,\mathrm dx}&=\int\frac {\frac{t^2+2}{2t^2}\,\mathrm dt}{\big(\frac{(t^2-2)^2}{4t^2}+1\big)\frac{t^2+2}{2t}}\\[1ex]
&=\int\frac{4t\,\mathrm dt}{t^4+4}\\[1ex]
&=2\int\frac{\mathrm d(\color{blue}{t^2})}{(\color{blue}{t^2})^2+4}\\[1ex]
&=\arctan\frac{t^2}2+C\\[1ex]
&=\arctan\frac{(x+\sqrt{x^2+2})^2}2+C.
\end{align*} 所以 \begin{align*}
\int\frac{\sqrt{x^2+2}}{x^2+1}\,\mathrm dx&=\int\frac{(x^2+1)+1}{(x^2+1)\sqrt{x^2+2}}\,\mathrm dx\\[1ex]
&=\int\frac1{\sqrt{x^2+2}}\,\mathrm dx+\int\frac1{(x^2+1)\sqrt{x^2+2}}\,\mathrm dx\\[1ex]
&=\ln\big(x+\sqrt{x^2+2}\big)+\arctan\frac{(x+\sqrt{x^2+2})^2}2+C.
\end{align*} |
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Czhang271828
发表于 2022-12-6 13:00
