求问微积分题目,急
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x>0时,纯中胡
f(x)=x-ln(x+1),
f'(x)=1-1/(x+1)=x/(x+1)>0,做拦
f(x)>f(0)=0,x>ln(x+1)
同理g(x)=(x+1)ln(x+1)-x,
g'(x)=ln(x+1)+1-1>0,
g(x)>g(0)=0,ln(x+1)>培世x/(x+1)
f(x)=x-ln(x+1),
f'(x)=1-1/(x+1)=x/(x+1)>0,做拦
f(x)>f(0)=0,x>ln(x+1)
同理g(x)=(x+1)ln(x+1)-x,
g'(x)=ln(x+1)+1-1>0,
g(x)>g(0)=0,ln(x+1)>培世x/(x+1)
追答
=-∫ln(x+1)d1/(x+2)
=-ln(x+1)/(x+2)+∫1/(x+2)(x+1)dx
=-(ln2)/3+∫1/(x+1)-1/(x+2)dx
=-(ln2)/3+ln(x+1)-ln(x+2)
=(2/3)ln2-ln3-(-ln2)
=(5/3)ln2-ln3
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