Question

高数,不定积分,

Answer 1

要倒过来推：

∫dx/(x^2+1)=x/(x^2+1)-∫xd(1/(x^2+1))

=x/(x^2+1)+2∫x^2dx/(x^2+1)

=x/(x^2+1)+2∫(x^2+1-1)dx/(x^2+1)^2

=x/(x^2+1)+2∫dx/(x^2+1)-2∫dx/(x^2+1)^2

所以∫dx/(x^2+1)^2=x/(2(x^2+1))+1/2*∫dx/(x^2+1)