Question

求大神解答一下定积分的题目

求解答！

Answer 1

令 1+√x = u, 则 x = (u-1)^2, dx = 2(u-1)du
I = ∫<下1, 上3> [2(u-1)/u]du
= 2∫<下1, 上3> (1-1/u)du = 2[u-lnu]<下1, 上3>
= 2(3-ln3-1+0) = 2(2-ln3)

Answer 2

令x=t² dx=2t 2∫t/(1+t)dt=2(t-ln(1+t))+C