ln(1-根号X)dx的不定积分

题目

ln(1-根号X)dx的不定积分

答案

∫ln(1-√x) dx
= xln(1-√x) +(1/2)∫ √x/(1-√x) dx
= xln(1-√x) -(1/2) ∫ (1-√x -1)/(1-√x) dx
=xln(1-√x) -(1/2)x +(1/2)∫ 1/(1-√x) dx
let
x^(1/4) = sina
(1/4)x^(-3/4) dx = cosa da
dx = 4(sina)^3 cosa da
∫ 1/(1-√x) dx
=∫ [1/(cosa)^2] (4(sina)^3 cosa) da
= 4∫ (sina)^3/cosa da
= -4 ∫ [(1-(cosa)^2)/cosa ]dcosa
=-4[ ln|cosa|- (cosa)^2/2 ] + C'
=-4(ln|√(1-√x)| - (1-√x)/2 ) + C'
∫ln(1-√x) dx
=xln(1-√x) -(1/2)x +(1/2)∫ 1/(1-√x) dx
=xln(1-√x) -(1/2)x -2(ln|√(1-√x)| - (1-√x)/2 ) + C