Question

高数，积分，简单题，求详解

我知道应该把dx换成d√x，但是不知道具体该怎么换，求详解

Answer 1

令√x=t,x=t^2,dx=2tdt
x=1,t=1,x=4,t=2
原式=∫[1,2] arctantdt^2
=t^2arctant[1,2]-∫[1,2] t^2darctant
=4arctan4-arctan1-∫[1,2] t^2/(1+t^2)dt
=4arctan4-arctan1-∫[1,2] [1-1/(1+t^2)]dt
=4arctan4-arctan1-[t-arctant][1,2]
=4arctan4-arctan1-(2-arctan2-1+arctan1)
=4arctan4-1+arctan2

追问

答案似乎不对啊

追答

噢，代入值错误了
令√x=t,x=t^2,dx=2tdt
x=1,t=1,x=4,t=2
原式=∫[1,2] arctantdt^2
=t^2arctant[1,2]-∫[1,2] t^2darctant
=4arctan2-arctan1-∫[1,2] t^2/(1+t^2)dt
=4arctan2-arctan1-∫[1,2] [1-1/(1+t^2)]dt
=4arctan2-arctan1-[t-arctant][1,2]
=4arctan2-arctan1-(2-arctan2-1+arctan1)
=4arctan2-1+arctan2-2arctan1
=5arctan2-1-π/2