解
∫(x2−4x+11)2x2−3x+12dx
解
1471xarctan(71(x−2))+2871xsin(2arctan(71(x−2)))+14715arctan(71(x−2))+2871sin(2arctan(71(x−2)))−147(x2−4x+11)(x−2)(arctan(71(x−2))(x2−4x+11)+7(x−2))−2(x2−4x+11)1+C
解答ステップ
∫(x2−4x+11)2x2−3x+12dx
以下の部分分数を得る: (x2−4x+11)2x2−3x+12:x2−4x+111+(x2−4x+11)2x+1
=∫x2−4x+111+(x2−4x+11)2x+1dx
総和規則を適用する: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫x2−4x+111dx+∫(x2−4x+11)2x+1dx
∫x2−4x+111dx=71arctan(7x−2)
∫(x2−4x+11)2x+1dx=2871(2arctan(7x−2)+sin(2arctan(7x−2)))(x+1)−1417(x2−4x+11)(x−2)(arctan(71(x−2))(x2−4x+11)+7(x−2))+x2−4x+117
=71arctan(7x−2)+2871(2arctan(7x−2)+sin(2arctan(7x−2)))(x+1)−1417(x2−4x+11)(x−2)(arctan(71(x−2))(x2−4x+11)+7(x−2))+x2−4x+117
簡素化 71arctan(7x−2)+2871(2arctan(7x−2)+sin(2arctan(7x−2)))(x+1)−1417(x2−4x+11)(x−2)(arctan(71(x−2))(x2−4x+11)+7(x−2))+x2−4x+117:1471xarctan(71(x−2))+2871xsin(2arctan(71(x−2)))+14715arctan(71(x−2))+2871sin(2arctan(71(x−2)))−147(x2−4x+11)(x−2)(arctan(71(x−2))(x2−4x+11)+7(x−2))−2(x2−4x+11)1
=1471xarctan(71(x−2))+2871xsin(2arctan(71(x−2)))+14715arctan(71(x−2))+2871sin(2arctan(71(x−2)))−147(x2−4x+11)(x−2)(arctan(71(x−2))(x2−4x+11)+7(x−2))−2(x2−4x+11)1
定数を解答に追加する=1471xarctan(71(x−2))+2871xsin(2arctan(71(x−2)))+14715arctan(71(x−2))+2871sin(2arctan(71(x−2)))−147(x2−4x+11)(x−2)(arctan(71(x−2))(x2−4x+11)+7(x−2))−2(x2−4x+11)1+C