Question

35和40题求不定积分

Answer 1

换元法求解不定积分。

Answer 2

换元法求解不定积分。

Answer 3

(35)
∫dx/[1+√(1-x^2)]
let
x=sinu
dx=cosu du
∫dx/[1+√(1-x^2)]
=∫cosu/(1+cosu) du
=∫cosu.(1-cosu)/ (sinu)^2 du
=∫[cosu-(cosu)^2]/ (sinu)^2 du
=∫[cosu/(sinu)^2 du - ∫(cotu)^2 du
=∫dsinu/(sinu)^2 - ∫[(cscu)^2 -1 ] du
=-1/sinu + cotu + x + C
=-1/x + √(1-x^2)/x + x + C
(40)
∫dx/[√x+x^(1/4)]
let
u= x^(1/4)
du = (1/4)x^(-3/4) dx
dx = 4u^3 . du
--------
∫dx/[√x+x^(1/4)]
=∫4u^3 . du/(u^2+u)
=4∫u^2 /(u+1) du
=4∫u^2 /(u+1) du
=4∫[u-1 + 1/(u+1) ] du
= 4[ (1/2)u^2 - u + ln|u+1| ] +C
= 4[ (1/2)x^(1/2) - x^(1/4) + ln|x^(1/4)+1| ] +C

Answer 4

陆续更新，欢迎讨论~

追答

请题主采个纳吧~