利用换元积分法计算定积分?
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(1) 记 3+5x = u,则 x = (u-3)/5, dx = (1/5)du
∫<-1, 5> 2dx/(3+5x) = (2/5)∫<-2, 28> du/u
= (2/5)[ln|u|]<-2, 28> = (2/5)ln14
(2) 记 √x = u,则 x = u^2, dx = 2udu
∫<1, 4> √xdx/(1+√x) = 2∫<1, 2> u^2du/(1+u)
= 2∫<1, 2> (u^2-1+1)du/(1+u)
= 2∫<1, 2> [u-1+1/(1+u)]du
= [u^2 - 2u + 2ln(1+u)]<1, 2>
= 4-4+2ln3-1+2-2ln2 = 1+2ln(3/2)
∫<-1, 5> 2dx/(3+5x) = (2/5)∫<-2, 28> du/u
= (2/5)[ln|u|]<-2, 28> = (2/5)ln14
(2) 记 √x = u,则 x = u^2, dx = 2udu
∫<1, 4> √xdx/(1+√x) = 2∫<1, 2> u^2du/(1+u)
= 2∫<1, 2> (u^2-1+1)du/(1+u)
= 2∫<1, 2> [u-1+1/(1+u)]du
= [u^2 - 2u + 2ln(1+u)]<1, 2>
= 4-4+2ln3-1+2-2ln2 = 1+2ln(3/2)
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