∫(0到1) T²/(1+T²) dT
=∫(0到1) (1+T²-1)/(1+T²) dT
=∫(0到1) [1-1/(1+T²) dT
=[T-arctanT](0到1)
=(1-arctan1)-(0-arctan0)
=1-π/4
至于前面那个d/dx到底干什么的?
是求d/dx ∫(0到1) T²/(1+T²) dT?
由于定积分是个常数
∴原式=(1-π/4)*d(1)/dx
=(1-π/4)*0
=0