【例2】 求下列不定积分
(1)∫x2(x+1)3dx.(2)∫1+x2x4−x2dx.(3)∫1+sinx1−sinxdx.解
(1)∫x2(x+1)3dx(2)∫1+x2x4−x2dx(3)∫1+sinx1−sinxdx=∫x2x3+3x2+3x+1dx=∫(x+3+x3+x21)dx=21x2+3x+3ln∣x∣−x1+C.=∫1+x2(x4−1)−(1+x2)+2dx=∫(x2−1−1+1+x22)dx=31x3−2x+2arctanx+C.=∫cos2x(1−sinx)2dx=∫(sec2x−2secx⋅tanx+tan2x)dx=tanx−2secx+tanx−x+C=2tanx−2secx−x+C.
