求不定积分∫1/x(1-x^¼)dx

Question

推导详细些，谢谢了

Answer 1

解：∫1/x(1-x^¼)dx
令：x^1/4=t 则， x=t^4
所以， dx=4t^3dt
所以，∫1/x(1-x^¼)dx=∫4t^3/t^4(1-t)dt=4∫1/t(1-t)dt=4∫[1/t+1/(1-t)]dt [说明：1/t+1/(1-t)=1/t(1-t)]
=4[lnt-ln(1-t)]+C=4lnt/(1-t)+C=4lnx^1/4/(1-x^1/4)+C
=lnx/(1-x^1/4)^4+C

Answer 2

您好，这道题用换元法，令t=x^¼，∫1/x(1-x^¼)dx=∫1/t^4(1-t)dt^4=∫4/t(1-t)dt=4∫[1/t+1/(1-t)]dt=4lnt-4ln(1-t)+C=lnx-4ln(1-x^¼)+C

Answer 3

∫ 1/[x(1 - x^(1/4))] dx

= ∫ 1/[x^(1/4) • x^(3/4) • (1 - x^(1/4))] dx
= ∫ 1/[x^(1/4) • (1 - x^(1/4))] d[4x^(1/4)]
= 4∫ [(1 - x^(1/4)) + x^(1/4)]/[x^(1/4) • (1 - x^(1/4))] d[x^(1/4)]
= 4∫ [1/x^(1/4) + 1/(1 - x^(1/4))] d[x^(1/4)]
= 4ln|x^(1/4)| + (- 4)ln|1 - x^(1/4)| + C
= ln|x| - 4ln|1 - x^(1/4)| + C

换元其实也比较麻烦，倒不如做的干净利落点吧。。

Answer 4

∫1/[x(1-x^¼)]dx

let
x^(1/8) = siny
(1/8)x^(-7/8) dx = cosydy
dx = 8(siny)^7cosy dy

∫1/[x(1-x^¼)] dx
=∫(1/[ (siny)^8 (cosy)^2] )(8(siny)^7cosy dy)
=8∫ 1/(sinycosy) dy
=16∫ 1/sin2y dy
=8ln|csc2y-cot2y| + C