计算∫[-1,0][(3x^4+3x^2+1)\1+x^2]dx

∫[-1,0][(3x^4+3x^2+1)/(1+x^2)]dx
=∫[-1,0] 3x^2dx +∫[-1,0] 1/(1+x^2)]dx
=x^3 | [-1,0] + acrtanx | [-1,0]
=1+π/21是怎么算出来的？画图吗？原函数代上下限x^3 | [-1,0] = 0^3 - (-1)^3 = 1