第一题答案是：10/3 - 2√3
第二题答案是：4 - 2arctan(2)
∫(x-1)/√(1+x) dx,from 0 to 2
=∫(1+x-2)/√(1+x) dx
=∫[(1+x)/√(1+x) - 2/√(1+x)] dx
=∫[√(1+x) - 2/√(1+x)] d(1+x)
=(2/3)(1+x)^(3/2) - 2*2√(1+x)
=[(2/3)(1+2)^(3/2)-4√(1+2)] - [(2/3)(1+0)^(3/2)-4√(1+0)]
=-2√3 - (2/3-4)
=10/3 - 2√3
∫√(x-1) / x dx,from 1 to 5
Let：u=√(x-1),x=u²+1,dx=2u du
When x=1,u=0、when x=5,u=2
Then the integral becomes
2∫u²/(1+u²) du,from 0 to 2
=2∫(1+u²-1)/(1+u²) du
=2∫[1 - 1/(1+u²)] du
=2u - 2arctan(u)
=[2(2) - 2arctan(2)] - [0 - 2arctan(0)]
=4 - 2arctan(2)