Find the integral from 1 to 16 for (x-1)/(x+x^0.5)dx?
回答 (2)
2020-06-08 11:37 pm
Let u = √x
u(1) = 1
u(16) = 4
du = dx/(2√x) = dx/(2u)
dx = 2udu
∫[(x-1)/(x + √x)]dx
= 2∫[(u²-1)/(u² + u)]udu
= 2∫[(u+1)(u-1)u/(u² + u)]du
= 2∫[(u+1)(u-1)/(u+1)]du
= 2∫(u-1)du
= u² - 2u
= u(u-2) : u = 1 to 4
= 8 + 1
= 9
u(1) = 1
u(16) = 4
du = dx/(2√x) = dx/(2u)
dx = 2udu
∫[(x-1)/(x + √x)]dx
= 2∫[(u²-1)/(u² + u)]udu
= 2∫[(u+1)(u-1)u/(u² + u)]du
= 2∫[(u+1)(u-1)/(u+1)]du
= 2∫(u-1)du
= u² - 2u
= u(u-2) : u = 1 to 4
= 8 + 1
= 9
收錄日期: 2021-04-24 07:50:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200608151010AA59dZx