解答
∫((x−2)(x2−2x+2)2)(x−1)dx
解答
41ln∣x−2∣−81(lnx2−2x+2+2arctan(x−1))+4(x2−2x+2)2x4arctan(x−1)+x4sin(2arctan(x−1))−x3−4x3arctan(x−1)−4x3sin(2arctan(x−1))+8x2arctan(x−1)+4x2+8x2sin(2arctan(x−1))−8xarctan(x−1)−6x−8xsin(2arctan(x−1))+4arctan(x−1)+4sin(2arctan(x−1))+4+C
求解步骤
∫(x−2)(x2−2x+2)2x−1dx
将(x−2)(x2−2x+2)2x−1用部份分式展开:4(x−2)1−4(x2−2x+2)x+2(x2−2x+2)2−x+2
=∫4(x−2)1−4(x2−2x+2)x+2(x2−2x+2)2−x+2dx
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫4(x−2)1dx−∫4(x2−2x+2)xdx+∫2(x2−2x+2)2−x+2dx
∫4(x−2)1dx=41ln∣x−2∣
∫4(x2−2x+2)xdx=81(lnx2−2x+2+2arctan(x−1))
∫2(x2−2x+2)2−x+2dx=4(x2−2x+2)x2−2x+2x3(x−1)+x2arctan(x−1)−x2−x2−2x+2x2(x−1)−x2−2x+22x2(x−1)+x2sin(2arctan(x−1))−2xarctan(x−1)+x+x2−2x+24x(x−1)−2xsin(2arctan(x−1))−x2−2x+22(x−1)+2arctan(x−1)+2sin(2arctan(x−1))+1
=41ln∣x−2∣−81(lnx2−2x+2+2arctan(x−1))+4(x2−2x+2)x2−2x+2x3(x−1)+x2arctan(x−1)−x2−x2−2x+2x2(x−1)−x2−2x+22x2(x−1)+x2sin(2arctan(x−1))−2xarctan(x−1)+x+x2−2x+24x(x−1)−2xsin(2arctan(x−1))−x2−2x+22(x−1)+2arctan(x−1)+2sin(2arctan(x−1))+1
4(x2−2x+2)x2−2x+2x3(x−1)+x2arctan(x−1)−x2−x2−2x+2x2(x−1)−x2−2x+22x2(x−1)+x2sin(2arctan(x−1))−2xarctan(x−1)+x+x2−2x+24x(x−1)−2xsin(2arctan(x−1))−x2−2x+22(x−1)+2arctan(x−1)+2sin(2arctan(x−1))+1=4(x2−2x+2)2x4arctan(x−1)+x4sin(2arctan(x−1))−x3−4x3arctan(x−1)−4x3sin(2arctan(x−1))+8x2arctan(x−1)+4x2+8x2sin(2arctan(x−1))−8xarctan(x−1)−6x−8xsin(2arctan(x−1))+4arctan(x−1)+4sin(2arctan(x−1))+4
=41ln∣x−2∣−81(lnx2−2x+2+2arctan(x−1))+4(x2−2x+2)2x4arctan(x−1)+x4sin(2arctan(x−1))−x3−4x3arctan(x−1)−4x3sin(2arctan(x−1))+8x2arctan(x−1)+4x2+8x2sin(2arctan(x−1))−8xarctan(x−1)−6x−8xsin(2arctan(x−1))+4arctan(x−1)+4sin(2arctan(x−1))+4
解答补常数=41ln∣x−2∣−81(lnx2−2x+2+2arctan(x−1))+4(x2−2x+2)2x4arctan(x−1)+x4sin(2arctan(x−1))−x3−4x3arctan(x−1)−4x3sin(2arctan(x−1))+8x2arctan(x−1)+4x2+8x2sin(2arctan(x−1))−8xarctan(x−1)−6x−8xsin(2arctan(x−1))+4arctan(x−1)+4sin(2arctan(x−1))+4+C