令t=x²+1，dt=2xdx，
则不定积分=1/2∫lntdt
=(tlnt-t)/2+C
=(x²+1)ln(x²+1)/2-x²/2+C
则不定积分=1/2∫lntdt
=(tlnt-t)/2+C
=(x²+1)ln(x²+1)/2-x²/2+C
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第1个回答 2018-01-03
∫xln(x² +1）dx
=(1/2)∫ln(x² +1）dx^2
=(1/2)x^2ln(x^2+1)-∫x^2*2x/(1+x^2)dx
=(1/2)x^2ln(x^2+1)-2∫[(x^2+1)x-x]/(1+x^2)dx
=(1/2)x^2ln(x^2+1)-2∫xdx+2∫x/(1+x^2)dx
=(1/2)x^2ln(x^2+1)-x^2+ln(1+x^2)+c追问
=(1/2)∫ln(x² +1）dx^2
=(1/2)x^2ln(x^2+1)-∫x^2*2x/(1+x^2)dx
=(1/2)x^2ln(x^2+1)-2∫[(x^2+1)x-x]/(1+x^2)dx
=(1/2)x^2ln(x^2+1)-2∫xdx+2∫x/(1+x^2)dx
=(1/2)x^2ln(x^2+1)-x^2+ln(1+x^2)+c追问
谢谢
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