Question

麻烦各位高手帮忙解答一下，求过程。谢谢啦！

Answer 1

1 > u = x^2 > 0,

du = 2xdx.

原式=(1/2)S(1-u)^(1/2)du/(1+u)^(1/2)

1 > t = [(1-u)/(1+u)]^(1/2) > 0. 1> u > 0.

t^2 = (1-u)/(1+u),
t^2 + t^2u = 1 - u,
(1+t^2)u = 1 - t^2,
u = (1-t^2)/(1+t^2),
du = [-2t/(1+t^2) - 2t(1-t^2)/(1+t^2)^2]dt

原式=(1/2)St[-2t/(1+t^2) - 2t(1-t^2)/(1+t^2)^2]dt
= St^2dt/(1+t^2) - St^2(1-t^2)/(1+t^2)^2
= S(1+t^2-1)dt/(1+t^2) - S(1+t^2-1)(1-t^2)dt/(1+t^2)^2
= Sdt - Sdt/(1+t^2) - S(1-t^2)dt/(1+t^2) + S(1-t^2)dt/(1+t^2)^2
= t - arctan(t) - S(2-1-t^2)dt/(1+t^2) + S(2-1-t^2)dt/(1+t^2)^2
= t - arctan(t) - 2Sdt/(1+t^2) + Sdt + 2Sdt/(1+t^2)^2 - Sdt/(1+t^2)
= t - arctan(t) - 2arctan(t) + t + 2Sdt/(1+t^2)^2 - arctan(t)
= 2t - 4arctan(t) + 2Sdt/(1+t^2)^2

t = tan(y), 0< t<1, 0< y< PI/4.
y = arctan(t).
dt = [sec(y)]^2dy,

原式=2t - 4arctan(t) + 2S[sec(y)]^2dy/[sec(y)]^4
= 2t - 4arctan(t) + 2Sdy/[sec(y)]^2
= 2t - 4arctan(t) + 2S[cos(y)]^2dy
= 2t - 4arctan(t) + S[1+cos(2y)]dy
= 2t - 4arctan(t) + Sdy + (1/2)Sd[sin(2y)]
= 2t - 4arctan(t) + y + (1/2)sin(2y) + C
= 2t - 4arctan(t) + arctan(t) + (1/2)sin[2arctan(t)] + C
= 2[(1-x^2)/(1+x^2)]^(1/2) - 4arctan{[(1-x^2)/(1+x^2)]^(1/2)} + (1/2)sin{ 2arctan{[(1-x^2)/(1+x^2)]^(1/2)} } + C,

其中，C为任意常数。