Delpiero23
幼苗
共回答了33个问题 举报
∫(2x+1)/(x²+4x+3)dx
=∫(x+x+1)/(x+1)(x+3)dx
=∫[x/(x+1)(x+3)+1/(x+3)]dx
=∫[x/(x²+4x+3)]dx+ln|x+3|+c
=(1/2)ln|x²+4x+3|-(3/2)∫1/(x+1)(x+3)dx+ln|x+3|+c
=(1/2)ln|x²+4x+3|-(3/4)∫[1/(x+1)-1/(x+3)]dx+ln|x+3|+c
=(1/2)ln|x²+4x+3|-(3/4)ln|(x+1)/(x+3)|+ln|x+3|+c
1年前
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