解
∫(x+2)(x2+2x+4)21dx
解
161ln∣x+2∣+161(−21lnx2+2x+4+31arctan(31(x+1)))+41−1231x(2arctan(3x+1)+sin(2arctan(3x+1)))+613(x2+2x+4)(x+1)(arctan(31(x+1))(x2+2x+4)+3(x+1))+x2+2x+43+C
解答ステップ
∫(x+2)(x2+2x+4)21dx
平方完成する x2+2x+4:(x+1)2+3
=∫(x+2)((x+1)2+3)21dx
u置換積分法を適用する
=∫(u+1)(u2+3)21du
以下の部分分数を得る: (u+1)(u2+3)21:16(u+1)1+16(u2+3)−u+1+4(u2+3)2−u+1
=∫16(u+1)1+16(u2+3)−u+1+4(u2+3)2−u+1du
総和規則を適用する: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫16(u+1)1du+∫16(u2+3)−u+1du+∫4(u2+3)2−u+1du
∫16(u+1)1du=161ln∣u+1∣
∫16(u2+3)−u+1du=161(−21lnu2+3+31arctan(3u))
∫4(u2+3)2−u+1du=41631(−u+1)(arctan(3u)+21sin(2arctan(3u)))+613(3+u2)u(arctan(3u)(3+u2)+3u)+u2+33
=161ln∣u+1∣+161(−21lnu2+3+31arctan(3u))+41631(−u+1)(arctan(3u)+21sin(2arctan(3u)))+613(3+u2)u(arctan(3u)(3+u2)+3u)+u2+33
代用を戻す u=x+1=161ln∣x+1+1∣+161(−21ln(x+1)2+3+31arctan(3x+1))+41631(−(x+1)+1)(arctan(3x+1)+21sin(2arctan(3x+1)))+613(3+(x+1)2)(x+1)(arctan(3x+1)(3+(x+1)2)+3(x+1))+(x+1)2+33
簡素化 161ln∣x+1+1∣+161(−21ln(x+1)2+3+31arctan(3x+1))+41631(−(x+1)+1)(arctan(3x+1)+21sin(2arctan(3x+1)))+613(3+(x+1)2)(x+1)(arctan(3x+1)(3+(x+1)2)+3(x+1))+(x+1)2+33:161ln∣x+2∣+161(−21lnx2+2x+4+31arctan(31(x+1)))+41−1231x(2arctan(3x+1)+sin(2arctan(3x+1)))+613(x2+2x+4)(x+1)(arctan(31(x+1))(x2+2x+4)+3(x+1))+x2+2x+43
=161ln∣x+2∣+161(−21lnx2+2x+4+31arctan(31(x+1)))+41−1231x(2arctan(3x+1)+sin(2arctan(3x+1)))+613(x2+2x+4)(x+1)(arctan(31(x+1))(x2+2x+4)+3(x+1))+x2+2x+43
定数を解答に追加する=161ln∣x+2∣+161(−21lnx2+2x+4+31arctan(31(x+1)))+41−1231x(2arctan(3x+1)+sin(2arctan(3x+1)))+613(x2+2x+4)(x+1)(arctan(31(x+1))(x2+2x+4)+3(x+1))+x2+2x+43+C