解
∫(x2−4x+10)2x2−3x+11dx
解
1261xarctan(61(x−2))+2461xsin(2arctan(61(x−2)))+12613arctan(61(x−2))+2461sin(2arctan(61(x−2)))−126(x2−4x+10)(x−2)(arctan(61(x−2))(x2−4x+10)+6(x−2))−2(x2−4x+10)1+C
解答ステップ
∫(x2−4x+10)2x2−3x+11dx
以下の部分分数を得る: (x2−4x+10)2x2−3x+11:x2−4x+101+(x2−4x+10)2x+1
=∫x2−4x+101+(x2−4x+10)2x+1dx
総和規則を適用する: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫x2−4x+101dx+∫(x2−4x+10)2x+1dx
∫x2−4x+101dx=61arctan(6x−2)
∫(x2−4x+10)2x+1dx=2461(2arctan(6x−2)+sin(2arctan(6x−2)))(x+1)−1216(x2−4x+10)(x−2)(arctan(61(x−2))(x2−4x+10)+6(x−2))+x2−4x+106
=61arctan(6x−2)+2461(2arctan(6x−2)+sin(2arctan(6x−2)))(x+1)−1216(x2−4x+10)(x−2)(arctan(61(x−2))(x2−4x+10)+6(x−2))+x2−4x+106
簡素化 61arctan(6x−2)+2461(2arctan(6x−2)+sin(2arctan(6x−2)))(x+1)−1216(x2−4x+10)(x−2)(arctan(61(x−2))(x2−4x+10)+6(x−2))+x2−4x+106:1261xarctan(61(x−2))+2461xsin(2arctan(61(x−2)))+12613arctan(61(x−2))+2461sin(2arctan(61(x−2)))−126(x2−4x+10)(x−2)(arctan(61(x−2))(x2−4x+10)+6(x−2))−2(x2−4x+10)1
=1261xarctan(61(x−2))+2461xsin(2arctan(61(x−2)))+12613arctan(61(x−2))+2461sin(2arctan(61(x−2)))−126(x2−4x+10)(x−2)(arctan(61(x−2))(x2−4x+10)+6(x−2))−2(x2−4x+10)1
定数を解答に追加する=1261xarctan(61(x−2))+2461xsin(2arctan(61(x−2)))+12613arctan(61(x−2))+2461sin(2arctan(61(x−2)))−126(x2−4x+10)(x−2)(arctan(61(x−2))(x2−4x+10)+6(x−2))−2(x2−4x+10)1+C