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#2 21. 12. 2016 01:26
Re: Sum of 3 definite Integral
![$
\left(\frac{x^2-x}{x^3-3x+1}\right)^2&=\left(\frac{Ax^2+Bx+C}{x^3-3x+1}\right)'+\frac{Dx^2+Ex+F}{x^3-3x+1}\\[.5\baselineskip]
\left(\frac{x^2-x}{x^3-3x+1}\right)^2&=\frac{(2Ax+B)(x^3-3x+1)-(Ax^2+Bx+C)(3x^2-3)}{(x^3-3x+1)^2}+\frac{Dx^2+Ex+F}{x^3-3x+1}
$](/mathtex/b6/b6d3201179dc741da321b49a08a6e7b6.gif "kopírovat do textarea")
![$
\int\left(\frac{x^2-x}{x^3-3x+1}\right)^2\,\mathrm dx
&=-\frac 19\cdot\frac{5x^2-8x+2}{x^3-3x+1}
+\frac 29\int\frac{2x-1}{x^3-3x+1}\,\mathrm dx\\[.5\baselineskip]
\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)\left(\frac{x^2-x}{x^3-3x+1}\right)^2\,\mathrm dx
&=\left.-\frac 19\cdot\frac{5x^2-8x+2}{x^3-3x+1}\right|_{-5}^{-2}\left.-\frac 19\cdot\frac{5x^2-8x+2}{x^3-3x+1}\right|_{\frac 16}^{\frac 13}\left.-\frac 19\cdot\frac{5x^2-8x+2}{x^3-3x+1}\right|_{\frac 65}^{\frac 32}\\[.5\baselineskip]
&\quad+\frac 29\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)\frac{2x-1}{x^3-3x+1}\,\mathrm dx\\[.5\baselineskip]
&=\frac{624}{109}+\frac 29\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)\frac{2x-1}{x^3-3x+1}\,\mathrm dx
$](/mathtex/00/0031ff32a511b30c60fb85df222d0a44.gif "kopírovat do textarea")
are mutually distinct real numbers then the general rational function
, i.e.![$
I=\int\frac{2x-1}{x^3-3x+1}\,\mathrm dx
&=\underbrace{\frac{4\cos\frac{2\pi}9-1}{4(\cos\frac{2\pi}9-\cos\frac{4\pi}9)(\cos\frac{2\pi}9-\cos\frac{8\pi}9)}}_{\mathsf F_1}\log\left|x-2\cos\frac{2\pi}9\right|\\[.5\baselineskip]
&\quad+\underbrace{\frac{4\cos\frac{4\pi}9-1}{4(\cos\frac{4\pi}9-\cos\frac{2\pi}9)(\cos\frac{4\pi}9-\cos\frac{8\pi}9)}}_{\mathsf F_2}\log\left|x-\cos\frac{4\pi}9\right|\\[.5\baselineskip]
&\quad+\underbrace{\frac{4\cos\frac{8\pi}9-1}{4(\cos\frac{8\pi}9-\cos\frac{2\pi}9)(\cos\frac{8\pi}9-\cos\frac{4\pi}9)}}_{\mathsf F_3}\log\left|x-\cos\frac{8\pi}9\right|
$](/mathtex/4e/4e3ed806d2c50915cc08c15f495a1db7.gif "kopírovat do textarea")
, 
, 
can be reduced substantially. We just apply the identities
![$
\mathsf F_1
&=\frac{4\cos\frac{2\pi}9-1}{4(\cos\frac{2\pi}9-\cos\frac{4\pi}9)(\cos\frac{2\pi}9-\cos\frac{8\pi}9)}
=\frac{4\cos\frac{2\pi}9-1}{16\sin\frac{\pi}3\sin\frac{\pi}9\sin\frac{\pi}3\sin\frac{5\pi}9}
=\frac{4\cos\frac{2\pi}9-1}{12\sin\frac{\pi}9\sin\frac{5\pi}9}
=\frac{4\cos\frac{2\pi}9-1}{-6(\cos\frac{2\pi}3-\cos\frac{4\pi}9)}\\[.5\baselineskip]
&=\frac 13\cdot\frac{4\cos\frac{2\pi}9-1}{1+2(2\cos^2\frac{2\pi}9-1)}
=\frac 13\cdot\frac{4\cos\frac{2\pi}9-1}{4\cos^2\frac{2\pi}9-1}
=\frac 13\cdot\frac{\cos\frac{2\pi}9(4\cos\frac{2\pi}9-1)}{4\cos^3\frac{2\pi}9-\cos\frac{2\pi}9}
=\frac 13\cdot\frac{\cos\frac{2\pi}9(4\cos\frac{2\pi}9-1)}{\cos\frac{2\pi}3+3\cos\frac{2\pi}9-\cos\frac{2\pi}9}\\[.5\baselineskip]
&=\frac 13\cdot\frac{\cos\frac{2\pi}9(4\cos\frac{2\pi}9-1)}{-\frac 12+2\cos\frac{2\pi}9}
=\frac 23\cdot\frac{\cos\frac{2\pi}9(4\cos\frac{2\pi}9-1)}{4\cos\frac{2\pi}9-1}
=\boldsymbol{\frac 23\cos\frac{2\pi}9}\\[\baselineskip]
\mathsf F_2
&=\frac{4\cos\frac{4\pi}9-1}{4(\cos\frac{4\pi}9-\cos\frac{2\pi}9)(\cos\frac{4\pi}9-\cos\frac{8\pi}9)}
=\frac{4\cos\frac{4\pi}9-1}{-16\sin\frac{\pi}3\sin\frac{\pi}9\sin\frac{2\pi}3\sin\frac{2\pi}9}
=\frac{4\cos\frac{4\pi}9-1}{-12\sin\frac{8\pi}9\sin\frac{2\pi}9}
=\frac{4\cos\frac{4\pi}9-1}{6(\cos\frac{10\pi}9-\cos\frac{2\pi}3)}\\[.5\baselineskip]
&=\frac{4\cos\frac{4\pi}9-1}{6(\cos\frac{8\pi}9+\frac 12)}
=\frac 13\cdot\frac{4\cos\frac{4\pi}9-1}{2(2\cos^2\frac{4\pi}9-1)+1}
=\frac 13\cdot\frac{4\cos\frac{4\pi}9-1}{4\cos^2\frac{4\pi}9-1}
=\frac 13\cdot\frac{\cos\frac{4\pi}9(4\cos\frac{4\pi}9-1)}{4\cos^3\frac{4\pi}9-\cos\frac{4\pi}9}\\[.5\baselineskip]
&=\frac 13\cdot\frac{\cos\frac{4\pi}9(4\cos\frac{4\pi}9-1)}{\cos\frac{4\pi}3+3\cos\frac{4\pi}9-\cos\frac{4\pi}9}
=\frac 13\cdot\frac{\cos\frac{4\pi}9(4\cos\frac{4\pi}9-1)}{-\frac 12+2\cos\frac{4\pi}9}
=\frac 23\cdot\frac{\cos\frac{4\pi}9(4\cos\frac{4\pi}9-1)}{4\cos\frac{4\pi}9-1}
=\boldsymbol{\frac 23\cos\frac{4\pi}9}\\[\baselineskip]
\mathsf F_3
&=\frac{4\cos\frac{8\pi}9-1}{4(\cos\frac{8\pi}9-\cos\frac{2\pi}9)(\cos\frac{8\pi}9-\cos\frac{4\pi}9)}
=\frac{4\cos\frac{8\pi}9-1}{16\sin\frac{\pi}3\sin\frac{5\pi}9\sin\frac{\pi}3\sin\frac{2\pi}9}
=\frac{4\cos\frac{8\pi}9-1}{12\sin\frac{5\pi}9\sin\frac{2\pi}9}
=\frac{4\cos\frac{8\pi}9-1}{-6(\cos\frac{7\pi}9-\cos\frac{\pi}3)}\\[.5\baselineskip]
&=\frac{4\cos\frac{8\pi}9-1}{6(\cos\frac{16\pi}9+\frac 12)}
=\frac 13\cdot\frac{4\cos\frac{8\pi}9-1}{2(2\cos^2\frac{8\pi}9-1)+1}
=\frac 13\cdot\frac{4\cos\frac{8\pi}9-1}{4\cos^2\frac{8\pi}9-1}
=\frac 13\cdot\frac{\cos\frac{8\pi}9(4\cos\frac{8\pi}9-1)}{4\cos^3\frac{8\pi}9-\cos\frac{8\pi}9}\\[.5\baselineskip]
&=\frac 13\cdot\frac{\cos\frac{8\pi}9(4\cos\frac{8\pi}9-1)}{\cos\frac{8\pi}3+3\cos\frac{8\pi}9-\cos\frac{8\pi}9}
=\frac 13\cdot\frac{\cos\frac{8\pi}9(4\cos\frac{8\pi}9-1)}{-\frac 12+2\cos\frac{8\pi}9}
=\frac 23\cdot\frac{\cos\frac{8\pi}9(4\cos\frac{8\pi}9-1)}{4\cos\frac{8\pi}9-1}
=\boldsymbol{\frac 23\cos\frac{8\pi}9}
$](/mathtex/04/0447d189159ba3d864affb28b56436d8.gif "kopírovat do textarea")
![$
\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)&\frac{2x-1}{x^3-3x+1}\,\mathrm dx\\[.5\baselineskip]
&=\left.\frac 23\left(\cos\frac{2\pi}9\log\left|x-2\cos\frac{2\pi}9\right|
+\cos\frac{4\pi}9\log\left|x-\cos\frac{4\pi}9\right|
+\cos\frac{8\pi}9\log\left|x-\cos\frac{8\pi}9\right|\right)\right|_{-5}^{-2}\\[.5\baselineskip]
&\quad+\left.\frac 23\left(\cos\frac{2\pi}9\log\left|x-2\cos\frac{2\pi}9\right|
+\cos\frac{4\pi}9\log\left|x-\cos\frac{4\pi}9\right|
+\cos\frac{8\pi}9\log\left|x-\cos\frac{8\pi}9\right|\right)\right|_{\frac 16}^{\frac 13}\\[.5\baselineskip]
&\quad+\left.\frac 23\left(\cos\frac{2\pi}9\log\left|x-2\cos\frac{2\pi}9\right|
+\cos\frac{4\pi}9\log\left|x-\cos\frac{4\pi}9\right|
+\cos\frac{8\pi}9\log\left|x-\cos\frac{8\pi}9\right|\right)\right|_{\frac 65}^{\frac 32}\\[.5\baselineskip]
&=\frac 23\cos\frac{2\pi}9\log\Biggl|\underbrace{\frac{(-2-2\cos\frac{2\pi}9)(\frac 13-2\cos\frac{2\pi}9)(\frac 32-2\cos\frac{2\pi}9)}{(-5-2\cos\frac{2\pi}9)(\frac 16-2\cos\frac{2\pi}9)(\frac 65-2\cos\frac{2\pi}9)}}_{\mathsf F_4}\Biggr|\\[.5\baselineskip]
&\quad+\frac 23\cos\frac{4\pi}9\log\Biggl|\underbrace{\frac{(-2-2\cos\frac{4\pi}9)(\frac 13-2\cos\frac{4\pi}9)(\frac 32-2\cos\frac{4\pi}9)}{(-5-2\cos\frac{4\pi}9)(\frac 16-2\cos\frac{4\pi}9)(\frac 65-2\cos\frac{4\pi}9)}}_{\mathsf F_5}\Biggr|\\[.5\baselineskip]
&\quad+\frac 23\cos\frac{8\pi}9\log\Biggl|\underbrace{\frac{(-2-2\cos\frac{8\pi}9)(\frac 13-2\cos\frac{8\pi}9)(\frac 32-2\cos\frac{8\pi}9)}{(-5-2\cos\frac{8\pi}9)(\frac 16-2\cos\frac{8\pi}9)(\frac 65-2\cos\frac{8\pi}9)}}_{\mathsf F_6}\Biggr|
$](/mathtex/23/230233d3bd3ef84e5ed72d7266231a7e.gif "kopírovat do textarea")
, 
, 
can be simplified using the trigonometric identities 
, 
. We rearrange ![$
\mathsf F_4
&=\frac{(-2-2\cos\frac{2\pi}9)(\frac 13-2\cos\frac{2\pi}9)(\frac 32-2\cos\frac{2\pi}9)}{(-5-2\cos\frac{2\pi}9)(\frac 16-2\cos\frac{2\pi}9)(\frac 65-2\cos\frac{2\pi}9)}
=5\cdot\frac{24\cos^3\frac{2\pi}9+2\cos^2\frac{2\pi}9-19\cos\frac{2\pi}9+3}{120\cos^3\frac{2\pi}9+218\cos^2\frac{2\pi}9-199\cos\frac{2\pi}9+15}\\[.5\baselineskip]
&=5\cdot\frac{6(\cos\frac{2\pi}3+3\cos\frac{2\pi}9)+\cos\frac{4\pi}9+1-19\cos\frac{2\pi}9+3}{30(\cos\frac{2\pi}3+3\cos\frac{2\pi}9)+109(\cos\frac{4\pi}9+1)-199\cos\frac{2\pi}9+15}\\[.5\baselineskip]
&=5\cdot\frac{1-\cos\frac{2\pi}9+\cos\frac{4\pi}9}{109-109\cos\frac{2\pi}9+109\cos\frac{4\pi}9}
=\frac 5{109}\cdot\frac{1-\cos\frac{2\pi}9+\cos\frac{4\pi}9}{1-\cos\frac{2\pi}9+\cos\frac{4\pi}9}
=\boldsymbol{\frac 5{109}}\\[\baselineskip]
\mathsf F_5
&=\frac{(-2-2\cos\frac{4\pi}9)(\frac 13-2\cos\frac{4\pi}9)(\frac 32-2\cos\frac{4\pi}9)}{(-5-2\cos\frac{4\pi}9)(\frac 16-2\cos\frac{4\pi}9)(\frac 65-2\cos\frac{4\pi}9)}=\dots=\boldsymbol{\frac 5{109}}\\[\baselineskip]
\mathsf F_6
&=\frac{(-2-2\cos\frac{8\pi}9)(\frac 13-2\cos\frac{8\pi}9)(\frac 32-2\cos\frac{8\pi}9)}{(-5-2\cos\frac{8\pi}9)(\frac 16-2\cos\frac{8\pi}9)(\frac 65-2\cos\frac{8\pi}9)}=\dots=\boldsymbol{\frac 5{109}}\\[\baselineskip]
$](/mathtex/c1/c14f97ede70f3c6603270f3b01c01638.gif "kopírovat do textarea")
![$
\boldsymbol{\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)\frac{2x-1}{x^3-3x+1}\,\mathrm dx}
&=\frac 23\cos\frac{2\pi}9\log\frac 5{109}+\frac 23\cos\frac{4\pi}9\log\frac 5{109}+\frac 23\cos\frac{8\pi}9\log\frac 5{109}\\[.5\baselineskip]
&=\frac 23\log\frac 5{109}\left(\cos\frac{2\pi}9+\cos\frac{4\pi}9+\cos\frac{8\pi}9\right)\\[.5\baselineskip]
&=\frac 23\log\frac 5{109}\left(2\cos\frac{\pi}3\cos\frac{\pi}9-\cos\frac{\pi}9\right)\\[.5\baselineskip]
&=\frac 23\log\frac 5{109}\left(\cos\frac{\pi}9-\cos\frac{\pi}9\right)=\boldsymbol 0
$](/mathtex/54/54396b4783c6d36e415ba3f8b227b759.gif "kopírovat do textarea")
![$
\large\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)\left(\frac{x^2-x}{x^3-3x+1}\right)^2\,\mathrm dx
&=\frac{624}{109}+\frac 29\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)\frac{2x-1}{x^3-3x+1}\,\mathrm dx\\[\baselineskip]
\boldsymbol{\color{blue}\left(\int_{-5}^{-2}+\int_{\frac{1}{6}}^{\frac{1}{3}}+\int_{\frac{6}{5}}^{\frac{3}{2}}\right)\left(\frac{x^2-x}{x^3-3x+1}\right)^2\,\mathrm dx}&\boldsymbol{\color{blue}=\frac{624}{109}}
$](/mathtex/26/26e65271f3e7f457091e222761afbb00.gif "kopírovat do textarea")
Backslash je v TeXu tak důležitý jako nekonečno při dělení nulou v tělesech charakteristiky 0.
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#3 08. 01. 2017 12:48 — Editoval stuart clark (08. 01. 2017 12:51)
- stuart clark
- Příspěvky: 1015
- Reputace: 7
, then 
and 
.
, then 
and 
.
