解答
积分 (x3−1)21
解答
−92ln∣x−1∣+91ln4x2+4x+4+932arctan(31(2x+1))+332arctan(31(2x+1))+932sin(2arctan(31(2x+1)))−9(x−1)1−6(x2+x+1)1−931(2arctan(31(2x+1))+sin(2arctan(31(2x+1))))+C
求解步骤
∫(x3−1)21dx
将(x3−1)21用部份分式展开:−9(x−1)2+9(x−1)21+9(x2+x+1)2x+3+3(x2+x+1)2x+1
=∫−9(x−1)2+9(x−1)21+9(x2+x+1)2x+3+3(x2+x+1)2x+1dx
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=−∫9(x−1)2dx+∫9(x−1)21dx+∫9(x2+x+1)2x+3dx+∫3(x2+x+1)2x+1dx
∫9(x−1)2dx=92ln∣x−1∣
∫9(x−1)21dx=−9(x−1)1
∫9(x2+x+1)2x+3dx=91(ln4x2+4x+4−32arctan(32x+1)+23arctan(31(2x+1)))
∫3(x2+x+1)2x+1dx=31(8(−4(4x2+4x+4)1−2431(2arctan(31(2x+1))+sin(2arctan(31(2x+1)))))+332(2arctan(32x+1)+sin(2arctan(32x+1))))
=−92ln∣x−1∣−9(x−1)1+91(ln4x2+4x+4−32arctan(32x+1)+23arctan(31(2x+1)))+31(8(−4(4x2+4x+4)1−2431(2arctan(31(2x+1))+sin(2arctan(31(2x+1)))))+332(2arctan(32x+1)+sin(2arctan(32x+1))))
化简 −92ln∣x−1∣−9(x−1)1+91(ln4x2+4x+4−32arctan(32x+1)+23arctan(31(2x+1)))+31(8(−4(4x2+4x+4)1−2431(2arctan(31(2x+1))+sin(2arctan(31(2x+1)))))+332(2arctan(32x+1)+sin(2arctan(32x+1)))):−92ln∣x−1∣+91ln4x2+4x+4+932arctan(31(2x+1))+332arctan(31(2x+1))+932sin(2arctan(31(2x+1)))−9(x−1)1−6(x2+x+1)1−931(2arctan(31(2x+1))+sin(2arctan(31(2x+1))))
=−92ln∣x−1∣+91ln4x2+4x+4+932arctan(31(2x+1))+332arctan(31(2x+1))+932sin(2arctan(31(2x+1)))−9(x−1)1−6(x2+x+1)1−931(2arctan(31(2x+1))+sin(2arctan(31(2x+1))))
解答补常数=−92ln∣x−1∣+91ln4x2+4x+4+932arctan(31(2x+1))+332arctan(31(2x+1))+932sin(2arctan(31(2x+1)))−9(x−1)1−6(x2+x+1)1−931(2arctan(31(2x+1))+sin(2arctan(31(2x+1))))+C