Question

用换元法怎么做

Answer 1

被积函数在[1,5]上连续,定积分存在.
令√(x-1)=t,x=t²+1,dx=2tdt.当x∈[1,5]时t∈[1,2]
原式=∫{1,2}2t²dt/(t²+1)
=2∫{1,2}(t²+1-1)dt/(t²+1)
=2∫{1,2}dt-2∫{1,2}dt/(t²+1)
=2(2-1)-2(arctan2-arctan1)
=2+π/2-2arctan2

Answer 2

let
x = (secu)^2
dx = 2(secu)^2.tanu du
x=1, u=0
x=5, u =arcsec(√5)
∫(1->5) √(x-1)/x dx
=∫(0->arcsec(√5)) (tanu / (secu)^2 ) ( 2(secu)^2.tanu du)
=2∫(0->arcsec(√5)) (tanu)^2 du

=2∫(0->arcsec(√5)) [(secu)^2-1] du
=2[ tanu - u ] |(0->arcsec(√5))
=2[ 2 - arcsec(√5) ]
=4 - 2arcsec(√5)