怎么用积分表达定积分的换元法?

如题所述

第1个回答  2024-01-13
∫xln(1+x)
dx,令u=x+1=∫(u-1)*lnu
du=∫ulnu
du-∫lnu
du=∫lnu
d(u²/2)-(ulnu-∫
du)=u²/2*lnu-∫u²/2
d(lnu)-ulnu+u=u²/2*lnu-1/2*∫u
du-ulnu+u=u²/2*lnu-1/4*u²-ulnu+u+C=(1/2)(x+1)²ln(x+1)-(1/4)(x+1)²-(x+1)ln(x+1)+x+1+C=(1/2)(x+1)²ln(x+1)-(1/4)(x+1)²-(x+1)ln(x+1)+x+C1
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