总觉得这种瑕积分还是先求出原函数比较方便些.
∫ xln(1 - x) dx
= ∫ ln(1 - x) d(x²/2)
= (x²/2)ln(1 - x) - (1/2)∫ x² * (- 1)/(1 - x) dx
= (x²/2)ln(1 - x) - (1/2)∫ x²/(x - 1) dx
= (x²/2)ln(1 - x) - (1/2)∫ [(x² - 1) + 1]/(x - 1) dx
= (x²/2)ln(1 - x) - (1/2)∫ [(x - 1)(x + 1) + 1]/(x - 1) dx
= (x²/2)ln(1 - x) - (1/2)∫ (x + 1) dx - (1/2)∫ dx/(x - 1)
= (x²/2)ln(1 - x) - (1/2)(x²/2 + x) - (1/2)ln|x - 1| + C
= (x²/2)ln(1 - x) - x²/4 - x/2 - (1/2)ln|x - 1| + C
= (1/2)(x² - 1)ln(1 - x) - (x/4)(x + 2) + C
∫(0→1) xln(1 - x) dx
= lim(x→1) [(1/2)(x² - 1)ln(1 - x) - (x/4)(x + 2)] - 0
= 0 - (1/4)(1 + 2)
= - 3/4