∫ x²/(1-x²)dx
=∫ (x²-1+1)/(1-x²)dx
=∫ (-1)dx+∫ 1/[(1-x)(1+x)] dx
=-x+0.5∫ 1/(1-x)dx+0.5∫ 1/(1+x)dx
=-x-0.5ln|1-x|+0.5ln|1+x|+C
=-x+0.5*ln|(1+x)/(1-x)|+C
求x^2/1-x^2dx的不定积分
∫ x²/(1-x²)dx
=∫ (x²-1+1)/(1-x²)dx
=∫ (-1)dx+∫ 1/[(1-x)(1+x)] dx
=-x+0.5∫ 1/(1-x)dx+0.5∫ 1/(1+x)dx
=-x-0.5ln|1-x|+0.5ln|1+x|+C
=-x+0.5*ln|(1+x)/(1-x)|+C