解答
积分 (x2+x+1)2x2
解答
332arctan(31(2x+1))+332arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−332sin(2arctan(31(2x+1)))+2(x2+x+1)1+C
求解步骤
∫(x2+x+1)2x2dx
将(x2+x+1)2x2用部份分式展开:x2+x+11+(x2+x+1)2−x−1
=∫x2+x+11+(x2+x+1)2−x−1dx
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫x2+x+11dx+∫(x2+x+1)2−x−1dx
∫x2+x+11dx=32arctan(32x+1)
∫(x2+x+1)2−x−1dx=−8(−4(4x2+4x+4)1−2431(2arctan(31(2x+1))+sin(2arctan(31(2x+1)))))−332(2arctan(32x+1)+sin(2arctan(32x+1)))
=32arctan(32x+1)−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 32arctan(32x+1)−8(−4(4x2+4x+4)1−2431(2arctan(31(2x+1))+sin(2arctan(31(2x+1)))))−332(2arctan(32x+1)+sin(2arctan(32x+1))):332arctan(31(2x+1))+332arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−332sin(2arctan(31(2x+1)))+2(x2+x+1)1
=332arctan(31(2x+1))+332arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−332sin(2arctan(31(2x+1)))+2(x2+x+1)1
解答补常数=332arctan(31(2x+1))+332arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−332sin(2arctan(31(2x+1)))+2(x2+x+1)1+C