Question

高数求导题，求大佬帮忙解答一下

Answer 1

Answer 2

let
x=tanu
dx=(secu)^2 du
∫dx/(x^2+1)^2
=∫(secu)^2 du/(secu)^4
=∫(cosu)^2 du
=(1/2) ∫(1+cos2u) du
=(1/2)[ u + (1/2)sin2u] +C
=(1/2)[ arctanx + x/(1+x^2) ] +C