Question

求解过程第五题

Answer 1

设√x＝t，范围是[1,2]
原式＝∫(1,2)2t^2/(1+t)dt
＝2∫(1,2)t-1+1/(t+1)dt
＝t^2-2t+2lnlt+1l
定积分结果为4-4+2ln3-(1-2+2ln2)
＝1+2ln(3/2)

Answer 2

=∫t/(1+t)dt²=∫(2t-2)+2/(t+1)dt=t²-2t+2ln(t+1)=x+2√x+2ln(√x+1)
=4+2*2+2ln3-(1+2+2ln2)=5+2ln1.5