Question

求定积分,上限为1,下限为0,求不定积分[x平方/(1+x平方）]dx

Answer 1

原式=∫[0,1]（1+x^2-1)dx/(1+x^2)
=∫[0,1]dx-∫[0,1]dx/(1+x^2)
=x[0,1]-arctanx[0,1]
=1-0-（π/4-0）
=1-π/4.