2道简单的定积分为方便书写,省略积分限以下2道都是一步做出来的,省略了中间步骤,1,∫ -u/根号(1+u^2)du= -根号(1+u^2)2,∫-2u/(1+u)=2u-2ln(1+u)第2题:∫-2udu/(u-1)=-2[u+ln(1-u)]1)∫-2udu/(u-1)=)-2∫u
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2道简单的定积分为方便书写,省略积分限以下2道都是一步做出来的,省略了中间步骤,1,∫ -u/根号(1+u^2)du= -根号(1+u^2)2,∫-2u/(1+u)=2u-2ln(1+u)第2题:∫-2udu/(u-1)=-2[u+ln(1-u)]1)∫-2udu/(u-1)=)-2∫u
2道简单的定积分
为方便书写,省略积分限
以下2道都是一步做出来的,省略了中间步骤,
1,∫ -u/根号(1+u^2)du= -根号(1+u^2)
2,∫-2u/(1+u)=2u-2ln(1+u)
第2题:∫-2udu/(u-1)=-2[u+ln(1-u)]
1)∫-2udu/(u-1)=)-2∫udu/(u-1)
=-2∫ud(u-1)/(u-1)
=-2∫[1+ 1/(u-1)]d(u-1)
=-2[u-1+ln(u-1)]
2) -2∫udu/(u-1)
=-2∫[1+ u/(u-1)]du
=-2[∫du +∫1/(u-1)d(u-1)]
=-2[u+ln(u-1)]
2道简单的定积分为方便书写,省略积分限以下2道都是一步做出来的,省略了中间步骤,1,∫ -u/根号(1+u^2)du= -根号(1+u^2)2,∫-2u/(1+u)=2u-2ln(1+u)第2题:∫-2udu/(u-1)=-2[u+ln(1-u)]1)∫-2udu/(u-1)=)-2∫u
1.∫ -u/根号(1+u^2)du
=-∫ u/根号(1+u^2)du
=-∫udu/根号(1+u^2) (利用d(1+u^2)=2udu)
=-∫d(1+u^2)/2根号(1+u^2)
t=1+u^2
=-∫dt/2根号t
=-根号t=-根号(1+u^2)
2.好像还有点问题~
∫-2udu/(u-1)=-2u-2ln(u-1)
-2u/(u-1)=-2-2/(u-1)
∫-2udu/(u-1)=∫[-2-2/(u-1)]du=∫-2du-
∫2/(u-1)d(u-1)=-2u-2ln(u-1)
第2题,两种做法答案不一
这没有问题,因为其实你算的就是不定积分,本来在后面都要加上一个常数C,C由初值条件确定,这在定积分会讲得,所以两种方法后面都要加上一个常数C,这就没有影响了~
定积分怎么没有积分上下限,就当作不定积分来做吧
解:1)因为d(sqrt(1+u^2))/du=u/sqrt(1+u^2)
所以∫-u/sqrt(1+u^2)du
=-∫u/sqrt(1+u^2)du
=-sqrt(1+u^2)+C
d(sqrt(1+u^2))/du=u/sqrt(1+u^2)这是常用的,要记住
2...
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定积分怎么没有积分上下限,就当作不定积分来做吧
解:1)因为d(sqrt(1+u^2))/du=u/sqrt(1+u^2)
所以∫-u/sqrt(1+u^2)du
=-∫u/sqrt(1+u^2)du
=-sqrt(1+u^2)+C
d(sqrt(1+u^2))/du=u/sqrt(1+u^2)这是常用的,要记住
2)
∫-2u/(u-1)du
=-2∫u/(u-1)du
=-2∫u/(u-1)d(u-1) 〔因为d(u-1)=du〕
=-2∫[(u-1)/(u-1)+1/(u-1)]d(u-1)
=-2∫d(u-1)-2∫1/(u-1)d(u-1)
=-2u-2ln(u-1)+C
两种答案是一样的
-2[u-1+ln(n-1)] +C
=-2u-2ln(u-1)+C+2
=-2u-2ln(u-1)+C1
也就是说C+2=C1,因为C+2和C1都可以看作任意常数,即使是定积分算出来也不影响答案,因为用牛顿莱布尼茨公式算的时候会代入积分上下限,这样会把常数减掉的
sqrt是根号的意思
收起
所以∫-u/sqrt(1+u^2)du
=-∫u/sqrt(1+u^2)du
=-sqrt(1+u^2)+C
d(sqrt(1+u^2))/du=u/sqrt(1+u^2)这是常用的,要记住
2)∫-2u/(1+u)du
=-2∫u/(1+u)du