如何用洛必达法则求导数

Question

如题所述

Answer 1

∫ xln(x+1)dx
= ∫ (x+1-1）ln(x+1)dx
= ∫ (x+1）ln(x+1)d(x+1) - ∫ ln(x+1)d(x+1)
= (1/2)(x+1)²{ln(x+1) - 1/2} - (x+1)ln(x+1) + (x+1) + c
= (1/2)(x+1)²ln(x+1) - 1/2} -(1/4)(x+1)² - (x+1)ln(x+1) + (x+1) + c
= (1/2)(x²-1) ln(x+1) - (1/4)(x²-2x-3) + c
= (1/2)(x²-1) ln(x+1) - (1/4)(x²-2x) + c