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∫[x/(x+3)]dx
=∫[(x+3-3)/(x+3)]dx
=∫[1-3/(x+3)]dx
=∫1dx-3∫[1/(x+3)]dx
=x-3ln(x+3)+C
所以原定积分=[x-3ln(x+3)]|<-1,-2>
=[(-2)-3ln1]-[(-1)-3ln2]
=[(-2)-0]+(1+3ln2)
=-1+3ln2
=∫[(x+3-3)/(x+3)]dx
=∫[1-3/(x+3)]dx
=∫1dx-3∫[1/(x+3)]dx
=x-3ln(x+3)+C
所以原定积分=[x-3ln(x+3)]|<-1,-2>
=[(-2)-3ln1]-[(-1)-3ln2]
=[(-2)-0]+(1+3ln2)
=-1+3ln2
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∫(-1->-2) x/(x+3) dx
=∫(-1->-2) [1 - 3/(x+3) ]dx
=[ x - 3ln|x+3|]|(-1->-2)
=( -1 - 3ln|-1+3| ) - ( -2 -3ln|-2+3| )
=-1-3ln2 +2
=1-3ln2
=∫(-1->-2) [1 - 3/(x+3) ]dx
=[ x - 3ln|x+3|]|(-1->-2)
=( -1 - 3ln|-1+3| ) - ( -2 -3ln|-2+3| )
=-1-3ln2 +2
=1-3ln2
追问
1-3?
追答
∫(-1->-2) x/(x+3) dx
=∫(-1->-2) {1 - [3/(x+3) ] }dx
=[ x - 3ln|x+3|]|(-1->-2)
= ( -2 -3ln|-2+3| )-( -1 - 3ln|-1+3| )
=-2-(-1-3ln2)
=-1+3ln2
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