Question

X*根号下9-X^2的不定积分

求大神帮我看看，写个详细的过程，拜托了





Answer 1

(1)
∫x√(9-x^2) dx
=(-1/2)∫√(9-x^2) d(9-x^2)
=(-1/3)(9-x^2)^(3/2) + C
(5)
let
u=xlnx
du =(1+lnx) dx
∫(1+lnx)/√(xlnx) dx
=∫du/√u
=2√u +C
=2√(xlnx) +C
(6)
let
u= lnx
du =(1/x) dx
∫√(1+lnx)/(xlnx) dx
=∫[√(1+u)/u ] du
let
u= (tanw)^2
du = 2(tanw). (secw)^2 dw
∫[√(1+u)/u ] du
=∫[ (secw)/ (tanw)^2 ] { 2(tanw). (secw)^2 dw }
=2∫[(secw)^3/ (tanw) ] dw
=2∫dw/ (sinw.(cosw)^2 ]
=2∫ sinwdw/ [(sinw)^2.(cosw)^2 ]
=2∫ dcosw/ [(sinw)^2.(cosw)^2 ]
=2∫ dcosw/ [(1-(cosw)^2) .(cosw)^2 ]
=2∫[1/(1-(cosw)^2) + 1/(cosw)^2 ] dcosw
=2[ arctan(cosw) - (1/cosw) ] +C
=2{ arctan[1/√(u+1)] - √(u+1) } +C
=2{ arctan[1/√(lnx+1)] - √(lnx+1) } +C
where

u= (tanw)^2
(secw)^2 = u+1
secw = √(u+1)
cosw =1/√(u+1)

追问

可以帮我看看第五题和第六题吗？有关于lnx的那题。感谢感谢

追答

(5)
let
u=xlnx
du =(1+lnx) dx
∫(1+lnx)/√(xlnx) dx
=∫du/√u
=2√u +C
=2√(xlnx) +C
(6)
let
u= lnx
du =(1/x) dx
∫√(1+lnx)/(xlnx) dx
=∫[√(1+u)/u ] du
let
u= (tanw)^2
du = 2(tanw). (secw)^2 dw
∫[√(1+u)/u ] du
=∫[ (secw)/ (tanw)^2 ] { 2(tanw). (secw)^2 dw }
=2∫[(secw)^3/ (tanw) ] dw
=2∫dw/ (sinw.(cosw)^2 ]
=2∫ sinwdw/ [(sinw)^2.(cosw)^2 ]
=-2∫ dcosw/ [(sinw)^2.(cosw)^2 ]
=-2∫ dcosw/ [(1-(cosw)^2) .(cosw)^2 ]
=-2∫[1/(1-(cosw)^2) + 1/(cosw)^2 ] dcosw
=-2[ arctan(cosw) - (1/cosw) ] +C
=-2{ arctan[1/√(u+1)] - √(u+1) } +C
=-2{ arctan[1/√(lnx+1)] - √(lnx+1) } +C
where

u= (tanw)^2
(secw)^2 = u+1
secw = √(u+1)
cosw =1/√(u+1)

追问

这是第六题的答案，有点儿不一样啊，大神