24/04/23 13:21:14.39 7Ack2Qhi.net
>>447
1 + √(1+x) = u,
とおくと
x = (u-1)^2 － 1,
dx = 2(u-1)du,
より
∫ √{1+√(1+x)} dx
= ∫ √u・2(u-1)du
= (4/5)u^{5/2} － (4/3)u^{3/2}
= (4/15)(3u－5)u^{3/2},
積分の範囲： 1+√2 ≦ u < 1+√5,
(与式) = (4/15){(13+√5)√(1+√5)－(4+√2)√(1+√2)}
= 5.0655498446