因为：f′（x）=11+(1+x1−x)2•(1−x)+(1+x)(1−x)2=11+x2=∞n＝0(−x2)n=∞n＝0(−1)nx2n，-1＜x＜1，所以：f（x）=∫x0f′(t)dt=∫x0 ∞n＝0(−1)nt2ndt=∞n＝0(−1)n∫x0t2ndt=∞n＝0(−1)n2n+1x2n+1，-1＜...