积分号,2,x(x+1)^1/2dx怎么算的?
问题描述:
积分号,2,x(x+1)^1/2dx怎么算的?
答
设t=根号(x+1)则x=t²-1原式=∫(1→根号3) 2(t²-1) t² dt=∫(1→根号3) 2t^4 - 2t^2 dt= (2/5)t^5 - (2/3)t^3 (1→根号3)= (18/5)根号3 - 2根号3 - 2/5 + 2/3=4/15 + (8/5)根号3...