Twitter\u306e\u30bf\u30a4\u30e0\u30e9\u30a4\u30f3\u3092\u773a\u3081\u3066\u3044\u305f\u3068\u3053\u308d\u3001$\\displaystyle\\int\\displaystyle\\frac{dx}{x^3+1}$\u3092\u8a08\u7b97\u3057\u3066\u3044\u305f\u30c4\u30a4\u30fc\u30c8\u304c\u76ee\u306b\u5165\u3063\u305f\u306e\u3067\u3001\u305d\u308c\u3092\u3061\u3087\u3044\u3068\u4e00\u822c\u5316\u3057\u3066\u3001
\n\\begin{align}
\nI &= \\displaystyle\\int\\displaystyle\\frac{dx}{x^3+a^3}\\label{eq:xcubeint}
\n\\end{align}
\n(\u305f\u3060\u3057\u3001$a \\gt 0$\u3068\u3059\u308b\u3002)\u3068\u3044\u3046\u5f0f\u3092\u8003\u3048\u3001\u8a08\u7b97\u3057\u3066\u307f\u308b\u3053\u3068\u306b\u3057\u307e\u3057\u305f\u3002<\/p>\n
\u6700\u521d\u306b(\\ref{eq:xcubeint})\u5f0f\u53f3\u8fba\u306e\u88ab\u7a4d\u5206\u95a2\u6570\u306b\u3064\u3044\u3066\u3001\u539f\u59cb\u95a2\u6570\u304c\u6c42\u3081\u3084\u3059\u3044\u5f62\u306b\u5909\u5f62\u3059\u308b\u3053\u3068\u3092\u8003\u3048\u307e\u3059\u3002<\/p>\n
(\\ref{eq:xcubeint})\u5f0f\u53f3\u8fba\u306e\u88ab\u7a4d\u5206\u95a2\u6570\u3092\u90e8\u5206\u5206\u6570\u5c55\u958b\u3059\u308b\u3068\u3001 (\\ref{eq:pfdsecond})\u5f0f\u306e\u53f3\u8fba\u306e\u62ec\u5f27\u5185\u306e\u5404\u9805\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u9805\u5225\u306b\u6c42\u3081\u307e\u3059\u3002<\/p>\n (\\ref{eq:pfdsecond})\u5f0f\u306e\u53f3\u8fba\u7b2c1\u9805\u306f\u3001 \u307e\u305f\u3001(\\ref{eq:pfdsecond})\u5f0f\u306e\u53f3\u8fba\u7b2c2\u9805\u306f\u3001 (\\ref{eq:pfdsecond})\u5f0f\u306e\u53f3\u8fba\u7b2c3\u9805\u306f\u3001\u7b2c1\u9805\u53ca\u3073\u7b2c2\u9805\u3068\u306f\u7570\u306a\u308b\u65b9\u6cd5\u3067\u8a08\u7b97\u3057\u307e\u3059\u3002 \u4e00\u65b9\u3001(\\ref{eq:xtheta})\u5f0f\u3092$\\theta$\u306b\u3064\u3044\u3066\u89e3\u304f\u3068\u3001 (\\ref{eq:pfdint_atfirst})\u3001(\\ref{eq:pfdint_atsecond})\u53ca\u3073(\\ref{eq:pfdint_atthird})\u306e\u5404\u5f0f\u3092(\\ref{eq:pfdsecond})\u5f0f\u306b\u4ee3\u5165\u3059\u308b\u3068\u2026 (\\ref{eq:fa})\u5f0f\u53ca\u3073(\\ref{eq:fasecond})\u5f0f\u306b$a=1$\u3092\u4ee3\u5165\u3059\u308b\u3068\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059\u3002 \u4e0d\u5b9a\u7a4d\u5206\u306e\u7d50\u679c\u306b\u306flog\u3084\u9006\u6b63\u63a5\u95a2\u6570(arctan)\u304c\u767b\u5834\u3059\u308b\u4e0a\u306b\u3001$f(x) = \\displaystyle\\frac{1}{x^3+a^3} (a \\gt 0)$\u306e\u30b0\u30e9\u30d5\u3082\u3044\u307e\u3044\u3061\u304d\u308c\u3044\u306a\u5f62\u3067\u306f\u306a\u3044\u305f\u3081\u306b\u3001\u9762\u7a4d\u3092\u8a08\u7b97\u3059\u308b\u6a5f\u4f1a\u3082\u3042\u307e\u308a\u306a\u3044\u3068\u3044\u3046\u3061\u3087\u3063\u3068\u4e0d\u9047\u306a\u611f\u3058\u3082\u3059\u308b\u95a2\u6570\u3067\u3059\u304c\u3001\u6642\u3005\u601d\u3044\u51fa\u3057\u3066\u3084\u3063\u3066\u3044\u305f\u3060\u3051\u308b\u3068\u5e78\u3044\u3067\u3059\u3002(\u00b4\u30fb\u03c9\u30fb\uff40)<\/p>\n \u3053\u306e\u8a18\u4e8b\u306f\u4ee5\u4e0a\u3067\u3059\u3002<\/p>\n","protected":false},"excerpt":{"rendered":" \u306f\u3058\u3081\u306b Twitter\u306e\u30bf\u30a4\u30e0\u30e9\u30a4\u30f3\u3092\u773a\u3081\u3066\u3044\u305f\u3068\u3053\u308d\u3001$\\displaystyle\\int\\displaystyle\\frac{dx}{x^3+1}$\u3092\u8a08\u7b97\u3057\u3066\u3044\u305f\u30c4\u30a4\u30fc\u30c8\u304c\u76ee\u306b\u5165\u3063\u305f\u306e\u3067\u3001\u305d\u308c\u3092\u3061\u3087\u3044\u3068\u4e00\u822c\u5316\u3057\u3066\u3001 \\begin{align} I &= \\displaystyle\\int\\displaystyle\\frac{dx}{x^3+a^3}\\label{eq:xcubein\u2026 Read More »<\/a><\/span><\/p>\n","protected":false},"author":1,"featured_media":5044,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13],"tags":[],"class_list":["post-5020","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-13"],"_links":{"self":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts\/5020","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5020"}],"version-history":[{"count":26,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts\/5020\/revisions"}],"predecessor-version":[{"id":9379,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts\/5020\/revisions\/9379"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/media\/5044"}],"wp:attachment":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5020"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5020"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\\begin{align}
\n\\frac{1}{x^3+a^3} &= \\frac{1}{3a^2}\\left( \\frac{1}{x+a} – \\frac{x-2a}{x^2-ax+a^2}\\right)\\label{eq:pfdfirst}
\n\\end{align}
\n\u3068\u306a\u308a\u307e\u3059(\u8a73\u7d30\u306b\u3064\u3044\u3066\u306f\u3053\u3061\u3089<\/a>\u53c2\u7167\u3002)\u304c\u3001\u3055\u3089\u306b(\\ref{eq:pfdfirst})\u5f0f\u306e\u53f3\u8fba\u7b2c2\u9805\u3092\u3001
\n\\begin{align}
\n\\frac{1}{x^3+a^3} &= \\frac{1}{3a^2}\\left( \\frac{1}{x+a} – \\frac{x-\\dfrac{a}{2}}{x^2-ax+a^2} + \\frac{3a}{2}\\frac{1}{x^2-ax+a^2}\\right) \\label{eq:pfdsecond}
\n\\end{align}
\n\u3068\u5909\u5f62\u3057\u307e\u3059\u3002<\/p>\n\u5404\u9805\u3054\u3068\u306b\u5909\u5f62\u3057\u307e\u3059\u3002<\/h2>\n
\u7b2c1\u9805\u53ca\u3073\u7b2c2\u9805<\/h3>\n
\n\\begin{align}
\n\\int \\frac{dx}{x+a} &= \\log\\left|x+a\\right|+C \\label{eq:pfdint_atfirst}
\n\\end{align}
\n\u3068\u306a\u308a\u307e\u3059($C$\u306f\u7a4d\u5206\u5b9a\u6570\u3067\u3059)\u3002<\/p>\n
\n\\begin{align}
\n{\\LARGE\\int}\\frac{x-\\dfrac{a}{2}}{x^2-ax+a^2}dx &= \\frac{1}{2}\\int\\frac{2x-a}{x^2-ax+a^2}dx \\nonumber\\cr
\n&= \\frac{1}{2}\\int\\frac{(x^2-ax+a^2)^{\\prime}}{x^2-ax+a^2}dx \\nonumber\\cr
\n&= \\frac{1}{2}\\log\\left|x^2-ax+a^2\\right|+C \\label{eq:pfdint_atsecond}
\n\\end{align}
\n\u3068\u306a\u308a\u307e\u3059\u3002<\/p>\n\u7b2c3\u9805<\/h3>\n
\n\\begin{align}
\n\\frac{1}{x^2-ax+a^2} &= \\frac{1}{\\left(x-\\dfrac{a}{2}\\right)^2+\\dfrac{3}{4}a^2} \\label{eq:square}
\n\\end{align}
\n\u3068\u5909\u5f62\u3067\u304d\u307e\u3059\u306e\u3067\u3001
\n\\begin{align}
\nx &= \\frac{\\sqrt{3}}{2}a\\tan\\theta + \\displaystyle\\frac{a}{2} \\label{eq:xtheta}
\n\\end{align}
\n\u3068\u304a\u304f\u3068\u3001$\\displaystyle\\frac{dx}{d\\theta} = \\displaystyle\\frac{\\sqrt{3}}{2}a\\cdot\\displaystyle\\frac{1}{\\cos^2\\theta}$\u3067\u3042\u308b\u3053\u3068\u304b\u3089\u3001
\n\\begin{align}
\n{\\Large\\int}\\frac{1}{\\left(x-\\dfrac{a}{2}\\right)^2+\\dfrac{3}{4}a^2}dx &= {\\Large\\int}\\frac{1}{\\left(\\dfrac{\\sqrt{3}}{2}a\\tan\\theta + \\dfrac{a}{2}-\\dfrac{a}{2}\\right)^2+\\dfrac{3}{4}a^2}\\dfrac{\\sqrt{3}}{2}a\\cdot\\frac{1}{\\cos^2\\theta}d\\theta \\nonumber\\cr
\n&= {\\Large\\int}\\frac{1}{\\left(\\dfrac{\\sqrt{3}}{2}a\\tan\\theta\\right)^2+\\dfrac{3}{4}a^2}\\frac{\\sqrt{3}}{2}a\\cdot\\frac{1}{\\cos^2\\theta}d\\theta \\nonumber\\cr
\n&= {\\Large\\int}\\frac{1}{\\dfrac{3}{4}a^2\\tan^2\\theta+\\dfrac{3}{4}a^2}\\frac{\\sqrt{3}}{2}a\\cdot\\frac{1}{\\cos^2\\theta}d\\theta \\nonumber\\cr
\n&= {\\Large\\int}\\frac{1}{\\dfrac{3}{4}a^2\\dfrac{1}{\\cos^2\\theta}}\\frac{\\sqrt{3}}{2}a\\cdot\\frac{1}{\\cos^2\\theta}d\\theta \\nonumber\\cr
\n&= \\int\\frac{2}{\\sqrt{3}a}d\\theta \\nonumber\\cr
\n&= \\frac{2}{\\sqrt{3}a}\\theta + C \\label{eq:third}
\n\\end{align}
\n\u306b\u306a\u308a\u307e\u3059\u3002<\/p>\n
\n\\begin{align}
\n\\theta &= \\arctan\\left[\\frac{2}{\\sqrt{3}a}\\left(x-\\frac{a}{2}\\right)\\right] \\label{eq:thetasolution}
\n\\end{align}
\n\u306b\u306a\u308a\u307e\u3059\u306e\u3067\u3001\u3053\u308c\u3092(\\ref{eq:third})\u5f0f\u306b\u4ee3\u5165\u3059\u308b\u3068\u3001
\n\\begin{align}
\n\\frac{2}{\\sqrt{3}a}\\theta + C &= \\frac{2}{\\sqrt{3}a}\\arctan\\left[\\frac{2}{\\sqrt{3}a}\\left(x-\\frac{a}{2}\\right)\\right] + C \\label{eq:pfdint_atthird}
\n\\end{align}
\n\u3068\u306a\u308a\u307e\u3059\u3002<\/p>\n\u6700\u7d42\u7d50\u679c<\/h2>\n
\n\\begin{align}
\n\\int\\displaystyle\\frac{dx}{x^3+a^3} &= \\frac{1}{3a^2}\\left[ \\log\\left|x+a\\right| – \\frac{1}{2}\\log\\left|x^2-ax+a^2\\right|\\right. \\nonumber\\cr
\n&{} + \\left.\\frac{3a}{2}\\cdot\\frac{2}{\\sqrt{3}a}\\arctan\\left[\\frac{2}{\\sqrt{3}a}\\left(x-\\frac{a}{2}\\right)\\right]\\right] + C \\nonumber\\cr
\n&= \\frac{1}{3a^2}\\left[ \\log\\left|x+a\\right| – \\frac{1}{2}\\log\\left|x^2-ax+a^2\\right|\\right. \\nonumber\\cr
\n&{} + \\left.\\sqrt{3}\\arctan\\left[\\frac{2}{\\sqrt{3}a}\\left(x-\\frac{a}{2}\\right)\\right]\\right] + C \\label{eq:fa}
\n\\end{align}
\n\u3068\u3044\u3046\u306a\u304b\u306a\u304b\u8907\u96d1\u306a\u5f0f\u306b\u306a\u308a\u307e\u3059\u304c\u3001$a \\gt 0$\u304b\u3064$x \\in {\\mathbb R}$\u3067\u3042\u308c\u3070\u3001(\\ref{eq:fa})\u5f0f\u306e\u53f3\u8fba\u7b2c2\u9805\u3092\u3061\u3087\u3044\u3068\u5909\u5f62\u3057\u3066\u7b2c1\u9805\u3068\u304f\u3063\u3064\u3051\u3066\u3001
\n\\begin{align}
\n\\int\\displaystyle\\frac{dx}{x^3+a^3} &= \\frac{1}{3a^2}\\left[ \\log\\left|\\frac{x+a}{\\sqrt{x^2-ax+a^2}}\\right| + \\sqrt{3}\\arctan\\left[\\frac{2}{\\sqrt{3}a}\\left(x-\\frac{a}{2}\\right)\\right]\\right] + C\\label{eq:fasecond}
\n\\end{align}
\n\u3068\u66f8\u3044\u3066\u3082\u7f70\u306f\u5f53\u305f\u3089\u306a\u3044\u306e\u3067\u306f\u306a\u3044\u304b\u3068\u601d\u3044\u307e\u3059\u3002$\\blacksquare$<\/p>\n$a = 1$\u306e\u5834\u5408<\/h2>\n
\n\\begin{align}
\n\\int\\displaystyle\\frac{dx}{x^3+1} &= \\frac{1}{3}\\left[ \\log\\left|x+1\\right| – \\frac{1}{2}\\log\\left|x^2-x+1\\right|+\\sqrt{3}\\arctan\\left[\\frac{2}{\\sqrt{3}}\\left(x-\\frac{1}{2}\\right)\\right]\\right] + C\\nonumber\\cr
\n&= \\frac{1}{3}\\left[ \\log\\left|\\frac{x+1}{\\sqrt{x^2-x+1}}\\right| + \\sqrt{3}\\arctan\\left[\\frac{2}{\\sqrt{3}}\\left(x-\\frac{1}{2}\\right)\\right]\\right] + C \\label{eq:aequalsone}
\n\\end{align}
\n\u3061\u3087\u3063\u3068\u3060\u3051\u7c21\u5358\u306a\u5f0f\u306b\u306a\u3063\u305f\u3093\u3058\u3083\u306a\u3044\u304b\u3068\u601d\u3044\u307e\u3059\u3002(\uff40\u30fb\u03c9\u30fb\u00b4)<\/p>\n\u307e\u3068\u3081<\/h2>\n