總覺得這種瑕積分還是先求出原函式比較方便些。

∫ 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