Question

不定积分问题

第六小题求解，谢谢

Answer 1

分享一种解法。∵√[(1-x)/(1+x)]=(1-x)/√(1-x²)，∴设x=sinα，
∴原式=∫(1-sinα)sinαdα。而，(1-sinα)sinα=sinα-sin²α=sinα+(cos2α-1)/2，
∴原式=-cosα-α/2+sin2α/4+C=(x/2)√(1-x²)-√(1-x²)-(1/2)arcsinx+C。
供参考。

Answer 2

右边等号的第二个等号就出现问题了。 1/√(x2-1)≠ -1/√(1-x2)

Answer 3

(6) 令 √[(1-x)/(1+x)] = u, 得 x = (1-u^2)/(1+u^2), dx = =4udu/(1+u^2)^2
I = ∫x√[(1-x)/(1+x)]dx = 4∫u^2(u^2-1)/(1+u^2)^3]du
= 4∫(u^4+2u^2+1-3u^2-3+2)/(1+u^2)^3]du
= 4∫[1/(1+u^2)-3/(1+u^2)^2+2/(1+u^2)^3]du
= arctanu - 12∫du/(1+u^2)^2 + 8∫du/(1+u^2)^3]
后两项令 u = tanv，
I2 = -12∫du/(1+u^2)^2 = -12∫dv/(secv)^2 = -12∫(cosv)^2dv
= -6∫(1+cos2v)dv = -6v - 3sin2v
I3 = 8∫du/(1+u^2)^3 = 8∫dv/(secv)^4 = 8∫(cosv)^4dv
= 2∫(1+cos2v)^2dv = 2∫[1+2cos2v+(cos2v)^2]dv
= ∫[3+4cos2v+cos4v]dv = 3v + 2sin2v + (1/4)sin4v
I2 + I3 = -3v - sin2v + (1/4)sin4v
= -3arctanu - 2u/(1+u^2) + u(1-u^2)/(1+u^2)^2
I = -2arctanu - 2u/(1+u^2) + u(1-u^2)/(1+u^2)^2 + C
= -2arctan√[(1-x)/(1+x)] - 2√[(1-x)/(1+x)]/[2/(1+x)]
+ √[(1-x)/(1+x)][2x/(1+x)]/[4/(1+x)^2] + C
= -2arctan√[(1-x)/(1+x)] - √(1-x^2) + (x/2)√(1-x^2) + C