求不定积分∫(ln^2)x/xdx用换元法解.

Question

求不定积分∫(ln^2)x/xdx
用换元法解.

Answer 1

(1)
∫xe^-x dx=-∫x d(e^-x)
=-xe^(-x)+∫e^-x dx
=-xe^(-x)-e^(-x)+C
=-(x+1)e^(-x)+C
(2)
∫x³lnx dx
=∫lnx d(x⁴/4)
=(1/4)x⁴lnx-(1/4)∫x⁴d(lnx)
=(1/4)x⁴lnx-(1/4)∫x³ dx
=(1/4)x⁴lnx-(1/4)*x⁴/4+C
=(1/4)x⁴lnx-(1/16)x⁴+C
=(1/16)x⁴(4lnx-1)+C
(3)
设u=x+1,x=u-1,dx=du
∫xln(x+1) dx
=∫(u-1)lnu du
=∫ulnu du-∫lnu du
=∫lnu d(u²/2)-(ulnu-∫u dlnu)
=(1/2)u²lnu-(1/2)∫u² dlnu-(ulnu-∫du)
=(1/2)u²lnu-(1/2)∫u du-(ulnu-1)
=(1/2)u²lnu-(1/4)u²-ulnu+C
=(1/2)(x+1)²ln(x+1)-(1/4)(x+1)²-(x+1)ln(x+1)+C
=(1/2)(x²-1)ln(x+1)-(1/4)x(x-2)+C