{"id":4234,"date":"2013-12-11T06:00:09","date_gmt":"2013-12-10T21:00:09","guid":{"rendered":"http:\/\/fujiitoshiki.com\/improvesociety\/?p=4234"},"modified":"2014-09-10T16:52:06","modified_gmt":"2014-09-10T07:52:06","slug":"post-4234","status":"publish","type":"post","link":"https:\/\/www.fujiitoshiki.com\/improvesociety\/?p=4234","title":{"rendered":"\u7a4d\u5206"},"content":{"rendered":"

\u7a4d\u5206<\/h3>\n

\u3000\u3042\u308b\u95a2\u6570\u306b\u3064\u3044\u3066 dy\/dx = f(x) \u3067\u3042\u308b\u6642\uff0cy \u306f\u6b21\u306e\u3088\u3046\u306b\u8868\u73fe\u3055\u308c\u307e\u3059\uff0e<\/p>\n

\\displaystyle \\int f(x)dx<\/p>\n

\u3000f(x) = \\frac{d}{dx}F(x) \u306e\u6642\uff0c\u7a4d\u5206\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u5b9a\u7fa9\u3055\u308c\u307e\u3059\uff0e<\/p>\n

\\displaystyle \\int_a^b f(x)dx = F(b) - F(a)<\/p>\n

\u7a4d\u5206\u516c\u5f0f<\/h3>\n

\u3000\u95a2\u6570 u, v \u304a\u3088\u3073\u5b9a\u6570 a, b, c, p \u306b\u3064\u3044\u3066\u4e0b\u8a18\u304c\u6210\u308a\u7acb\u3061\u307e\u3059\uff0e<\/p>\n

\\displaystyle \\int (u \\pm v)dx = \\int u dx \\pm \\int v dx\\\\\\vspace{0.2 in} \\int cu dx = c\\int u dx \\\\\\vspace{0.2 in} \\int u\\left(\\frac{dv}{dx}\\right) = uv - \\int v \\left(\\frac{du}{dx}\\right)dx \\\\\\vspace{0.2 in} \\int u dv = uv - \\int v du \\\\\\vspace{0.2 in} \\int F(u(x))dx = \\int F(w)\\frac{dw}{dw\/dx},\\ w = u(x)\n<\/p>\n

\\displaystyle \\int u^p du = \\frac{u^{p+1}}{p+1},\\ p \\neq -1\\\\\\vspace{0.2 in} \\int u^{-1}du = \\int \\frac{du}{u} = \\ln u\\\\\\vspace{0.2 in} \\int a^u du = \\frac{a^u}{\\ln a},\\ a \\neq 0,\\ 1\\\\\\vspace{0.2 in} \\int e^u du = e^u\n<\/p>\n

\\displaystyle \\int \\sin u\\ du = -\\cos{u}\\\\\\vspace{0.2 in} \\int \\cos u\\ du = \\sin{u}\\\\\\vspace{0.2 in} \\int \\tan u\\ du = -\\ln \\cos{u}\\\\\\vspace{0.2 in} \\int e^{au}\\sin{bu}\\ du = \\frac{e^{au}(a\\ \\sin{bu}- b\\ \\cos{bu})}{a^2 + b^2}\\\\\\vspace{0.2 in} \\int e^{au}\\cos{bu}\\ du = \\frac{e^{au}(a\\ \\cos{bu}+ b\\ \\sin{bu})}{a^2 + b^2}\\\\\\vspace{0.2 in} \\int \\frac{du}{\\sqrt{a^2 - u^2}} = \\sin^{-1}\\frac{u}{a}\\\\\\vspace{0.2 in} \\int \\frac{du}{u^2 + a^2} = \\frac{1}{a}\\tan^{-1}\\frac{u}{a}\\\\\\vspace{0.2 in} \\int \\frac{du}{\\sqrt{u^2 - a^2}} = \\ln(u + \\sqrt{u^2 - a^2})\\\\\\vspace{0.2 in} \\int \\frac{du}{\\sqrt{u^2 + a^2}} = \\ln(u + \\sqrt{u^2 + a^2})<\/p>\n