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本帖最后由 hbghlyj 于 2020-8-18 21:37 编辑 ①$\int\dfrac1{x^{2n+1}+1},n\in\mathbf{Z}$
$\dfrac1{x^{2n+1}+1}=\dfrac{1}{2n+1} \left(\dfrac{1}{x+1}+2\sum\limits_{k=1}^n \dfrac{x \cos \left(\dfrac{2k \pi }{2 n+1}\right)+1}{x^2+2 x \cos \left(\dfrac{2 k\pi }{2 n+1}\right)+1}\right)$
$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln\left(x^2+2 x \cos \left(\dfrac{2k \pi }{2 n+1}\right)+1\right)=\dfrac{2\left(x+\cos\left(\dfrac{2k\pi}{2n+1}\right)\right)}{x^2+2 x \cos \left(\dfrac{2k \pi }{2 n+1}\right)+1}$
$\int\dfrac1{x^2+2x\cos\left(\dfrac{2k\pi}{2n+1}\right)+1}=\dfrac1{\left(x+\cos\left(\dfrac{2k\pi}{2n+1}\right)\right)^2+\sin^2\dfrac{2k\pi}{2n+1}}=\dfrac1{\sin\dfrac{2k\pi}{2n+1}}\arctan\dfrac{x+\cos\left(\dfrac{2k\pi}{2n+1}\right)}{\sin{\dfrac{2k\pi}{2n+1}}}+C$
$\dfrac{x \cos \left(\dfrac{2 k\pi }{2 n+1}\right)+1}{x^2+2 x \cos \left(\dfrac{2k \pi }{2 n+1}\right)+1}=\dfrac{\cos\left(\dfrac{2k\pi}{2n+1}\right)}2\dfrac{2\left(x+\cos\left(\dfrac{2k\pi}{2n+1}\right)\right)}{x^2+2 x \cos \left(\dfrac{2k \pi }{2 n+1}\right)+1}+\dfrac{\sin^2\left(\dfrac{2k\pi}{2n+1}\right)}{x^2+2x\cos\left(\dfrac{2k\pi}{2n+1}\right)+1}$
$\therefore\int\dfrac1{x^{2n+1}+1}=\dfrac{1}{2n+1} \left(\ln(x+1)+2\sum\limits_{k=1}^n \left(\dfrac{\cos\left(\dfrac{2k\pi}{2n+1}\right)}2\ln\left(x^2+2 x \cos \left(\dfrac{2k \pi }{2 n+1}\right)+1\right)+\sin\dfrac{2k\pi}{2n+1}\arctan\dfrac{x+\cos\left(\dfrac{2k\pi}{2n+1}\right)}{\sin{\dfrac{2k\pi}{2n+1}}}\right)\right)+C$
②$\int\dfrac1{x^{2n}+1},n\in\mathbf{Z}$
$\dfrac1{x^{2n}+1}=\dfrac1n\sum\limits_{k=1}^{n}\dfrac{1-x\cos\dfrac{(2k-1)\pi}{2n}}{x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1}$
$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln\left(x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1\right)=\dfrac{2\left(x-\cos\dfrac{(2k-1)\pi}{2n}\right)}{x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1}$
$\int\dfrac1{x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1}\mathrm{d}x=\int\dfrac1{\left(x-\cos\dfrac{(2k-1)\pi}{2n}\right)^2+\sin^2\dfrac{(2k-1)\pi}{2n}}\mathrm{d}x=\dfrac1{\sin\dfrac{(2k-1)\pi}{2n}}\arctan\dfrac{x-\cos\dfrac{(2k-1)\pi}{2n}}{\sin\dfrac{(2k-1)\pi}{2n}}+C$
$\dfrac{1-x\cos\dfrac{(2k-1)\pi}{2n}}{x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1}=-\dfrac{\cos\dfrac{(2k-1)\pi}{2n}}2\dfrac{2\left(x-\cos\dfrac{(2k-1)\pi}{2n}\right)}{x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1}+\dfrac{\sin^2\dfrac{(2k-1)\pi}{2n}}{x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1}$
$\therefore\int\dfrac1{x^{2n}+1}=\frac1n\sum\limits_{k=1}^{n}\left(-\dfrac{\cos\dfrac{(2k-1)\pi}{2n}}2\ln\left(x^2-2x\cos\dfrac{(2k-1)\pi}{2n}+1\right)+\sin\dfrac{(2k-1)\pi}{2n}\arctan\dfrac{x-\cos\dfrac{(2k-1)\pi}{2n}}{\sin\dfrac{(2k-1)\pi}{2n}}\right)+C$ |
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