Question

求积分，求积分

Answer 1

原式=∫dt²/[1+(t²)²]
=arctan(t²)+C

Answer 2

let
t^2= tanu
2tdt = (secu)^2 du
∫ 2t/(1+t^4 ) dt
=∫ du
=u + C
=arctan(t^2) + C