如题所述
第1个回答 2017-03-16
∫ t^2/(1-t^2) dt
= -∫ dt + ∫ dt/(1-t^2)
= - t +∫ dt/(1-t^2)
let
t = siny
dt = cosy dy
∫ dt/(1-t^2)
=∫ secy dy
=ln|secy + tany |
=ln| (1+t)/√(1-t^2) |
=(1/2)ln|(1+t)/(1-t)|
∫ t^2/(1-t^2) dt
= - t +∫ dt/(1-t^2)
=-t +(1/2)ln|(1+t)/(1-t)| + C
= -∫ dt + ∫ dt/(1-t^2)
= - t +∫ dt/(1-t^2)
let
t = siny
dt = cosy dy
∫ dt/(1-t^2)
=∫ secy dy
=ln|secy + tany |
=ln| (1+t)/√(1-t^2) |
=(1/2)ln|(1+t)/(1-t)|
∫ t^2/(1-t^2) dt
= - t +∫ dt/(1-t^2)
=-t +(1/2)ln|(1+t)/(1-t)| + C
第2个回答 2017-03-16
∫(1-t²)²dt
=∫(t⁴-2t²+1)dt
=(1/5)t⁵-⅔t³+t +C
=(3t⁴-10t²+15)t/15 +C本回答被提问者和网友采纳
=∫(t⁴-2t²+1)dt
=(1/5)t⁵-⅔t³+t +C
=(3t⁴-10t²+15)t/15 +C本回答被提问者和网友采纳
