Question

如图求积分

Answer 1

用柯西-黎曼积分下的换元积分法，希望帮到你

Answer 2

令 √x = u， 则
I = ∫<2, 3> u(2udu)/(u-1) = 2∫<2, 3> (u^2-u+u-1+1)du/(u-1)
= 2∫<2, 3> [u+1+1/(u-1)]du
= 2[u^2/2+u+ln(u-1)]<2, 3> = 2(9/2+3+ln2-2-2) = 7+2ln2

Answer 3

x^2-x = (x-1/2)^2 -1/4
let
x-1/2 =(1/2)secu
dx= (1/2)secu.tanu du
x=4 , u = arcsec(7)
x=9, u =arcsec(17)
∫ (4->9) √x/√(x-1) dx
=∫ (4->9) x /√(x^2-x) dx
=∫ (arcsec7->arcsec17) { [(1/2)+(1/2)secu] /[(1/2)tanu] } . [ (1/2)secu.tanu du]
=(1/2)∫ (arcsec7->arcsec17) (1+secu)secu du
=(1/2) [ ln|secu+tanu| + tanu ]| (arcsec7->arcsec17)
=(1/2) [ ( ln|17+√288| + √288 ) - ( ln|7+√48| + √48 ) ]
=(1/2) [ ( ln|17+12√2| + 12√2 ) - ( ln|7+4√3| + 4√3 ) ]

Answer 4

运用换元法，希望对你有所帮助