Question

帮忙解一下这两题，谢谢

帮忙解一下这两题，谢谢不定积分

Answer 1

见图

Answer 2

(3)
∫ √x/(√x + 1) dx
let
y=√x
dy = dx/(2√x)
dx= 2y dy
∫ √x/(√x + 1) dx
=∫ 2y^2/(y + 1) dy
=∫ 2y^2/(y + 1) dy
=2∫ [ y- 1 + 1/(y + 1)] dy
= 2[ (1/2)y^2 - y + ln|y+1| ] + C
=2[ (1/2)x - √x + ln|√x+1| ] + C

(4)
let
x^(1/6) = tanu
(1/6)x^(-5/6) dx = (secu)^2 . du
dx = 6(tanu)^5 .(secu)^2 . du

∫dx/(1+ x^(1/3))

=∫6(tanu)^5 .(secu)^2 . du/ (secu)^2
=6∫(tanu)^5 . du
=6∫(tanu)^3 .[(secu)^2 -1] du
=(3/2) (tanu)^4 - 6∫(tanu)^3 du
=(3/2) (tanu)^4 - 6∫(tanu)[(secu)^2 -1] du
=(3/2) (tanu)^4 - 3(tanu)^2 + 6∫tanu du
=(3/2) (tanu)^4 - 3(tanu)^2 + 6ln|cosu| + C
=(3/2) x^(2/3) - 3x^(1/3) + 6ln|1/√(1+x^(1/3)| + C
=(3/2) x^(2/3) - 3x^(1/3) - 3ln|1+x^(1/3)| + C