Question

积分计算疑惑

Answer 1

设x=arctant,则t=tanx,dx=(secx)^2dx,
∫dt/(1+t^2)^2
=∫(secx)^2dx/(secx)^4
=∫(cosx)^2dx
=(1/2)∫(1+cos2x)dx
=(1/2)[x+(1/2)sin2x]+c
=(1/2)[arctant+t/(1+t^2)}+c.
可以吗？