诚邀Jzkyllcjl先生参与定量计算\(\int_2^∞\tfrac{1}{1+x^3}dx\)的值地讨论

春风晚霞

例：计算\(\int_2^∞\tfrac{1}{1+x^3}dx\)的值，结果要求精确到\(10^{-15}\).
【分析】
       关于\(\int_2^∞\tfrac{1}{1+x^3}dx\)精确到\(10^{-15}\)的计算，固然可用无穷级数展开求值，但春风晚霞认为用部分分式法直接积分更好。设\(\tfrac{1}{1+x^3}=\)\(\tfrac{A}{1+x}+\tfrac{Bx+c}{x^2-x+1}\)\(=\tfrac{(A+B)x^2+(B+C-A)x+(A+C)}{1+x^3}\)，即\(\tfrac{1}{1+x^3}\)\(=\tfrac{(A+B)x^2+(B+C-A)x+(A+C)}{1+x^3}\)，比较分子得：①、A+B=0；②、-A+B+C=0；③、A+C=1；联解方程①②③得：A=\(\tfrac{1}{3}\)；B=-\(\tfrac{1}{3}\)；C=\(\tfrac{2}{3}\)。所以\(\tfrac{1}{1+x^3}=\)\(\tfrac{1}{3}\times\tfrac{1}{1+x}-\)\(\tfrac{1}{3}\times\tfrac{x-2}{x^2-x+1}\)\(=\tfrac{1}{6}\times\tfrac{2}{1+x}-\)\(\tfrac{1}{6}\times\tfrac{2x-1}{x^2-x+1}+\)\(\tfrac{2}{(2x-1)^2+3}\)\(=\tfrac{1}{6}\)\((\tfrac{2}{1+x}+\tfrac{2x-1}{x^2-x+1}\))+\(\tfrac{1}{3}\)\(\tfrac{2}{(\tfrac{2x-1}{\sqrt 3})^2+1}\).
即\(\tfrac{1}{1+x^3}\)\(=\tfrac{1}{6}\)\((\tfrac{2}{1+x}+\tfrac{2x-1}{x^2-x+1}\))+\(\tfrac{1}{3}\)\(\tfrac{2}{(\tfrac{2x-1}{\sqrt 3})^2+1}\).
\begin{split}
&\qquad【解】根据分析\int_2^∞\tfrac{dx}{1+x^3}=\int_2^∞[\tfrac{1}{6}(\tfrac{2}{1+x}+\tfrac{2x-1}{x^2-x+1})+\tfrac{1}{3}\tfrac{2}{(\tfrac{2x-1}{\sqrt 3})^2+1}]dx)\\
&=\int_2^∞[\tfrac{1}{6}(\tfrac{2dx}{1+x}+\tfrac{d(x^2-x+1)}{x^2-x+1})+\tfrac{1}{\sqrt 3}\tfrac{d{(\tfrac{2x-1}{\sqrt 3})}}{(\tfrac{2x-1}{\sqrt 3})^2+1}]\\
&=[\tfrac{1}{6}Ln\tfrac{(x+1)^2}{x^2-x+1}+\tfrac{1}{\sqrt 3}arctan\tfrac{2x-1}{\sqrt 3}]|_2^∞\\
&=\displaystyle\lim_{x \to \infty}\tfrac{1}{6}Ln\tfrac{(x+1)^2}{x^2-x+1}+\displaystyle\lim_{x \to \infty}\tfrac{1}{\sqrt 3}arctan\tfrac{2x-1}{\sqrt 3}-\tfrac{1}{6}Ln\tfrac{(2+1)^2}{2^2-2+1}
-\tfrac{1}{\sqrt 3}arctan\tfrac{2\times2-1)}{\sqrt 3}\\
&=\tfrac{1}{6}Ln\displaystyle\lim_{x \to \infty}\tfrac{(x+1)^2}{x^2-x+1}+\tfrac{1}{\sqrt 3}arctan\displaystyle\lim_{x\to \infty}\tfrac{2x-1}{\sqrt 3}-\tfrac{1}{6}Ln\tfrac{9}{3}
-\tfrac{1}{\sqrt 3}arctan\tfrac{3}{\sqrt 3}\\
&=0+\tfrac{1}{\sqrt 3}\times\tfrac{π}{2}-\tfrac{1}{6}Ln3
-\tfrac{1}{\sqrt 3}arctan\sqrt 3\\
&=\tfrac{π}{2\sqrt 3}-\tfrac{1}{6}Ln3-\tfrac{π}{3\sqrt 3}=\tfrac{{\sqrt 3}π}{18}-\tfrac{1}{6}Ln3\approx 0.119197845927685.【解毕】
\end{split}

jzkyllcjl

你用了部分分式法 是对的，我支持你。

elim

jzkyllcjl 除了抄袭，一般不会解题．

jzkyllcjl

elim 发表于 2022-9-9 12:19
jzkyllcjl 除了抄袭，一般不会解题．

你只会抄袭错误的F（4），不会按照变上限积分自己计算原函数在4的数值。

elim

我不会算，吃狗屎啼猿声的jzkyllcjl 会算，算的怎样拿出来看看？