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源自知乎提问
题： $\int \frac {x^2+2}{(1+x+x^2)^2}\mathrm dx.$
尝试倒数代换
\begin{align*} &\quad \int \frac {x^2+2}{(1+x+x^2)^2}\mathrm dx\\[1em] {}\xlongequal{x{~}\mapsto {~}\frac 1t} &=\int \frac {\frac 1{t^2}+2}{\left(1+\frac 1t+\frac 1{t^2}\right)^2}\cdot \left(-\frac 1{t^2}\right)\mathrm dt\\[1em] &=-\int \frac {1+2t^2}{\left(t^2+t+1\right)^2}\mathrm dt\\[1em] &=-\int \frac {\color{blue}{2t^2+2t+2}-(2t+1)}{\left(t^2+t+1\right)^2}\mathrm dt\\[1em] &=-2\int \frac {\mathrm d\left(t+1/2\right)}{3/4+\left(t+1/2\right)^2}+\int \frac {\mathrm d{~}(t^2+t+1)}{\left(t^2+t+1\right)^2}\\[1em] &=-2\cdot \frac 1{\sqrt 3/2}\arctan \frac {t+1/2}{\sqrt 3/2}-\frac {1}{t^2+t+1}+C\\[1em] &=-\frac {x^2}{1+x+x^2}-\frac {4\sqrt 3}{3}\arctan\frac {2+x}{\sqrt 3x}+C. \end{align*}
PS：过程写起来是没问题的，只是这个结果与大家相差甚远~
后来，回应在评论区的质疑.
补充更新一下，仅说明当 $x\ne 0$ 时，上述结果是正确的，只与其它答主相差个常数 $\frac {4\sqrt 3\pi}9-1$.
由恒等式 $\arctan a+\arctan b=\arctan \frac {a+b}{1-ab}$ 知
\begin{align*} \arctan \frac {2+x}{\sqrt {3x}}+\arctan \sqrt 3&=\arctan \frac {\frac {2+x}{\sqrt {3x}}+\sqrt 3}{1-\frac {2+x}{\sqrt {3x}}\cdot \sqrt 3}\\[1em] &=\arctan\frac {2+x+3x}{\sqrt 3x-\sqrt 3(2+x)}\\[1em] &=\arctan\frac {2(2x+1)}{-2\sqrt 3}\\[1em] &=-\arctan \frac {2x+1}{\sqrt 3}, \end{align*}
即
$$\color{blue}{\boxed{\arctan \frac {2+x}{\sqrt {3x}}=-\arctan \frac {2x+1}{\sqrt 3}-\frac {\pi}3}},$$
从而
\begin{align*} &\quad \int \frac {x^2+2}{(1+x+x^2)^2}\mathrm dx\\[1em] &=-\frac {x^2}{1+x+x^2}-\frac {4\sqrt 3}{3}\arctan\frac {2+x}{\sqrt 3x}+C\\[1em] &=-\frac {x^2+x+1-(\color{blue}{x+1})}{x^2+x+1}-\frac {4\sqrt 3}{3}\cdot \left(\color{blue}{-\arctan \frac {2x+1}{\sqrt 3}-\frac {\pi}3}\right)+C\\[1em] &=\frac {x+1}{x^2+x+1}+\frac {4\sqrt 3}{3}\arctan \frac {2x+1}{\sqrt 3}+\color{red}{\frac {4\sqrt 3\pi}9-1}+C\\[1em] &=\frac {x+1}{x^2+x+1}+\frac {4\sqrt 3}{3}\arctan \frac {2x+1}{\sqrt 3}+\color{red}{C'} \end{align*}
此结果当 $x=0$ 时亦成立. |
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