Question

求33、34题解法

Answer 1

(33)
∫ √x/[ √x - x^(1/3) ] dx
let
u = x^(1/6)
du = (1/6)x^(-5/6) dx
dx = 6u^5 du

∫ √x/[ √x - x^(1/3) ] dx
=∫ { u^3/[ u^3 - u^2 ] } ( 6u^5 du )
=6∫ { u^6/ (u - 1) du
=6∫ [ u^5+u^4+u^3+u^2+u+1 + 1/ (u - 1) ] du
=6[ (1/6)u^6 + (1/5)u^5 +(1/4)u^4+(1/3)u^2+ u + ln|u-1| ] + C
=6[(1/6)x+(1/5)x^(5/6) +(1/4)x^(2/3)+(1/3)x^(1/3)+ x^(1/6) + ln|x^(1/6)-1|] + C

(34)
x^2-2x+5
=(x-1)^2 +4

let
x-1 = 2tanu
dx=2(secu)^2 . du

∫dx/ √(5-2x+x^2)
=∫2(secu)^2 . du/ [2secu]
=∫secu du
=ln|secu+ tanu| +C
=ln|(1/2)√(5-2x+x^2)+x-1| + C

Answer 2

33问=∫(1+lnx)/(xlnx)²dx=∫1/(xlnx)²d(xlnx)=-1/(xlnx)+C
34问=∫dx/√((x-1)²+4)=∫1/√(((x-1)/2)²+1)d((x-1)/2)，设tant=(x-1)/2，则=∫sectdt=ln|sect+tant|+C=ln|√(((x-1)/2)²+1)+(x-1)/2|+C=ln|√(x²-2x+5)+x-1|+C'