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一个不定积分的计算

热度 4已有 636 次阅读2017-4-3 11:15 | 不定积分

求不定积分
\[\int \frac{x^2dx}{(x^4+1)^2}.\]

解法1 注意到
\[x^2=\frac{1}{8}\left((x^2-\sqrt{2}x+1)^2-2(x^4+1)+(x^2+\sqrt{2}x+1)^2\right),\]
得到
\[\int\frac{x^2dx}{(x^4+1)^2}=\frac{1}{8}\int\frac{dx}{(x^2+\sqrt{2}x+1)^2}+\frac{1}{8}\int\frac{dx}{(x^2-\sqrt{2}x+1)^2}-\frac{1}{4}\int\frac{dx}{x^4+1}.\]
根据
\[\begin{array}{rl}\displaystyle\int\frac{dx}{x^2+\sqrt{2}x+1}&=\displaystyle\frac{x+\frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1}
+\int\frac{2x^2+2\sqrt{2}+1}{(x^2+\sqrt{2}+1)^2}dx\\
&=\displaystyle\frac{x+\frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1}+2\int\frac{dx}{x^2+\sqrt{2}x+1}-\int\frac{dx}{(x^2+\sqrt{2}x+1)^2},\end{array}\]
得到
\[\int\frac{dx}{(x^2+\sqrt{2}x+1)^2}=\frac{x+\frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1}+\int\frac{dx}{x^2+\sqrt{2}x+1}=\frac{x+\frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1}
+\sqrt{2}\arctan(\sqrt{2}x+1)+C.\] 同理
\[\int\frac{dx}{(x^2-\sqrt{2}x+1)^2}=\frac{x-\frac{\sqrt{2}}{2}}{x^2-\sqrt{2}x+1}+\int\frac{dx}{x^2-\sqrt{2}x+1}=\frac{x-\frac{\sqrt{2}}{2}}{x^2-\sqrt{2}x+1}
+\sqrt{2}\arctan(\sqrt{2}x-1)+C.\] 而
\[\int\frac{dx}{x^4+1}=\frac{1}{2}\int\frac{x^2+1}{x^4+1}dx-\frac{1}{2}\int\frac{x^2-1}{x^4+1}dx,\]
其中
\[\begin{array}{rl}\displaystyle\int\frac{x^2+1}{x^4+1}dx&=\displaystyle\frac{1}{2}\int\frac{x^2-\sqrt{2}x+1}{x^4+1}dx+\frac{1}{2}\int\frac{x^2+\sqrt{2}x+1}{x^4+1}dx\\
&=\displaystyle\frac{1}{2}\int\frac{dx}{x^2+\sqrt{2}x+1}+\frac{1}{2}\int\frac{dx}{x^2-\sqrt{2}x+1}\\
&=\displaystyle\frac{1}{\sqrt{2}}\arctan(\sqrt{2}x+1)+\frac{1}{\sqrt{2}}\arctan(\sqrt{2}x-1)+C,\end{array}\]
\[\int\frac{x^2-1}{x^4+1}dx=\int\frac{d(x+\frac{1}{x})}{(x+\frac{1}{x})^2-2}=\frac{1}{2\sqrt{2}}\ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}+C.\]

\[\int\frac{dx}{x^4+1}=\frac{1}{2\sqrt{2}}\arctan(\sqrt{2}x+1)+\frac{1}{2\sqrt{2}}\arctan(\sqrt{2}x-1)-
\frac{1}{4\sqrt{2}}\ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}+C.\]
最后得到
\[\begin{array}{rl}\displaystyle\int\frac{x^2dx}{(x^4+1)^2}&=\displaystyle\frac{1}{8\sqrt{2}}\frac{\sqrt{2}x+1}{x^2+\sqrt{2}x+1}
+\frac{1}{8\sqrt{2}}\arctan(\sqrt{2}x+1)\\
&+\displaystyle\frac{1}{8\sqrt{2}}\frac{\sqrt{2}x-1}{x^2-\sqrt{2}x+1}
+\frac{1}{8\sqrt{2}}\arctan(\sqrt{2}x-1)\\
&+\displaystyle\frac{1}{16\sqrt{2}}\ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}+C.\end{array}\]
解法2  先求$\int\frac{xdx}{(x^4+1)^2}$.
注意到\[\begin{split}\int
\frac{xdx}{x^4+1}=&\frac{1}{2}\int\frac{1}{x^4+1}d(x^2)\\
=&\frac{x^2}{2(x^4+1)}-\frac{1}{2}\int x^2d\left(\frac{1}{x^4+1}\right)\\
=&\frac{x^2}{2(x^4+1)}+2\int\frac{x^5dx}{(x^4+1)^2}\\
=&\frac{x^2}{2(x^4+1)}+2\int\frac{xdx}{x^4+1}-2\int\frac{xdx}{(x^4+1)^2},\end{split}\]
从而
\[\int\frac{xdx}{(x^4+1)^2}=\frac{x^2}{4(x^4+1)}+\frac{1}{2}\int\frac{xdx}{x^4+1}=
\frac{x^2}{4(x^4+1)}+\frac{1}{4}\arctan x^2+c.\] 因此
\[\begin{split}\int\frac{x^2dx}{(x^4+1)^2}=&\int x\cdot \frac{xdx}{(x^4+1)^2}\\
=& \frac{1}{4}\int xd\left(\frac{x^2}{x^4+1}+\arctan
x^2\right)\\
=&\frac{x^3}{4(x^4+1)}+\frac{1}{4}x\arctan
x^2-\frac{1}{4}\int\frac{x^2dx}{x^4+1}-\frac{1}{4}\int\arctan
x^2dx.\end{split}\] 注意
\[\int\arctan x^2dx=x\arctan x^2-2\int\frac{x^2dx}{x^4+1}.\]
因此
\[\int\frac{x^2dx}{(x^4+1)^2}=\frac{x^3}{4(x^4+1)}+\frac{1}{4}\int\frac{x^2dx}{x^4+1},\]
其中
\[\begin{split}\int\frac{x^2dx}{x^4+1}&=\frac{1}{2}\int\frac{(x^2+1)dx}{x^4+1}+\frac{1}{2}\int\frac{(x^2-1)dx}{x^4+1}\\
&=\frac{1}{4}\int\frac{dx}{x^2+\sqrt{2}x+1}+\frac{1}{4}\int\frac{dx}{x^2-\sqrt{2}x+1}
+\frac{1}{2}\int\frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-2}\\
&=\frac{1}{2\sqrt{2}}\arctan(\sqrt{2}x+1)+\frac{1}{2\sqrt{2}}\arctan(\sqrt{2}x-1)+\frac{1}{4\sqrt{2}}\ln
\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}+c.\end{split}\] 故
\[\begin{split}\int\frac{x^2dx}{(x^4+1)^2}
=&\frac{x^3}{4(x^4+1)}+\frac{1}{4}\int\frac{x^2dx}{x^4+1}\\
=&\frac{x^3}{4(x^4+1)}+\frac{1}{8\sqrt{2}}\arctan(\sqrt{2}x+1)+\frac{1}{8\sqrt{2}}\arctan(\sqrt{2}x-1)+\frac{1}{16\sqrt{2}}\ln
\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}+c.\end{split}\]



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