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I=1∫0(1x4+x2+1+x3(1−x2)3)dx
=1∫01x4+x2+1dx+1∫0x3(1−x2)3dx
=I1+I2
Trong đó
I2=1∫0x3(1−x2)3dx=[−x1010+3x810−x62+x44]10=140
I1=1∫01x4+x2+1dx=121∫0(x+1x2+x+1−x−1x2−x+1)dx
=141∫0(d(x2+x+1)x2+x+1−d(x2−x+1)x2−x+1)+141∫0(1x2+x+1+1x2−x+1)dx
=[14ln|x2+x+1x2−x+1|+2√3arctanx−1√3+2√3arctanx−1√3]10=π√312+ln24
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