24/04/26 18:12:29.72 oEIwRUvS.net
>>561
(3/2)∫ (√x)*ln(1+x) dx
= x^{3/2} ln(1+x) －∫ x^{3/2} /(x+1) dx (← 部分積分)
= x^{3/2} ln(1+x) －∫ {√x－1/√x + 1/((x+1)√x)} dx
= x^{3/2} ln(1+x) －(2/3)x^{3/2} + 2√x－2∫1/(x+1)･dx/(2√x)
= x^{3/2} ln(1+x) －(2/3)x^{3/2} + 2√x－2arctan(√x),
∵ x=uu とおくと
∫1/(x+1)・dx/(2√x) = ∫1/(uu+1) du = arctan(u) = arctan(√x)
x：0→1 として
(与式) = (2/3){ln(2) + 4/3－π/2} = 0.30379458…