Question

一道不定积分的求解

求不定积分∫ arctan√xdx详细解答~~~~~~

Answer 1

令√x = t，x = t²，dx = d(t²)
∫ arctan√x dx
= ∫ arctant d(t²)
= t²arctant - ∫ t²/(1 + t²) dt
= t²arctant - ∫ [(1 + t²) - 1]/(1 + t²) dt
= t²arctant - ∫ dt + ∫ dt/(1 + t²)
= t²arctant - t + arctant + C
= xarctan√x - √x + arctan√x + C

Answer 2

∫ arctan√xdx
=x arctan√x - (1/2)∫ √x/(1+x) dx
let
√x = tana
dx/(2√x)= (seca)^2da
dx = 2tana(seca)^2da
∫ √x/(1+x) dx
=∫ [tana/(seca)^2] (2tana(seca)^2)da
=2∫ (tana)^2 da
= 2∫ [(seca)^2 -1] da
=2(tana - a) + C'
=2(√x - arctan(√x) )+ C'
∫ arctan√xdx
=x arctan√x - (1/2)∫ √x/(1+x) dx
=x arctan√x -(√x - arctan(√x) ) + C

追问

请问第一部的分部积分 为什么是x arctan√x 我不明白哦

追答

∫ udv
=uv- ∫ vdu

∫ arctan√xdx

u= arctan√x , v= x
∫ arctan√xdx
=x arctan√x - (1/2)∫ √x/(1+x) dx