求x^2\(1-x^4)dx的积分

答案有错,
∫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)-1/(1+x^2)]dx
=1/2∫[1/(1-x^2)-1/(1+x^2)]dx
=1/2∫1/(1-x^2)dx-1/2∫1/(1+x^2)dx
=1/2∫1/2[1/(1-x)+1/(1+x)]dx-1/2∫1/(1+x^2)dx (*)
=1/2∫1/2*1/(1-x)dx+1/2∫1/2*1/(1+x)-1/2∫1/(1+x^2)dx
=-1/4∫1/(x-1)dx+1/4∫1/(x+1)dx-1/2∫1/(1+x^2)dx
=-1/4ln|x-1|+1/4ln|x+1|-1/2arctanx+c
=1/4(ln|x+1|-ln|x-1|)-1/2arctanx+c
=1/4ln(|x+1|/|x-1|)-1/2arctanx+c
=1/2∫1/2[1/(1-x)+1/(1+x)]dx-1/2∫1/(1+x^2)dx (*) 中若把第一个“＋”号错化为“－”号,就得到了错结果