求不定积分∫x^2/(1-x^4)dx

∫x^2/(1-x^4)dx=∫x^2/[(1-x^2)(1+x^2)]dx=1/2·∫1/(1-x^2)dx-1/2·∫1/(1+x^2)dx=1/4·∫1/(1-x)dx+1/4·∫1/(1+x)dx-1/2·∫1/(1+x^2)dx= -1/4·ln|1-x|+1/4·ln|1+x|-1/2·arctanx+C