Question

这三道题求解

Answer 1

(4)
∫dx/(1+sinx+cosx)
=∫(1-sinx-cosx)/(1-(sinx+cosx)^2 ) dx
=∫(1-sinx-cosx)/(-2sinx.cosx ) dx
=-(1/2) ∫ [1/(sinx.cosx) - 1/cosx - 1/sinx ] dx
=-(1/2) ∫ (2csc2x - secx - cscx ) dx
=-(1/2)[ln| csc2x-cot2x| -ln|secx+tanx| - ln|cscx-cotx| ] +C

(5)
∫dx/[1+(x+1)^(1/3)]

let
(x+1)^(1/3) = (tanu)^2
(1/3)(x+1)^(-2/3) dx = 2tanu. (secu)^2 du
dx =6 (tanu)^5 . (secu)^2 du

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

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

let
x^(1/4) = (tanu)^2
(1/4)x^(-3/4) dx = 2tanu. (secu)^2 du
dx = 8(tanu)^7. (secu)^2 du

∫dx/[x^(1/4)+1]
=8∫(tanu)^7 du
=8∫(tanu)^5 .[(secu)^2 -1] du
=(4/3)(tanu)^6 -8∫(tanu)^5 du
=(4/3)(tanu)^6 -8∫(tanu)^3.[(secu)^2 -1] du
=(4/3)(tanu)^6 -2(tanu)^4 +8∫(tanu)^3 du
=(4/3)(tanu)^6 -2(tanu)^4 +8∫(tanu).[(secu)^2 -1] du
=(4/3)(tanu)^6 -2(tanu)^4 +4(tanu)^2 -8∫tanu du
=(4/3)(tanu)^6 -2(tanu)^4 +4(tanu)^2 +8ln|cosu| + C'
=(4/3)x^(3/4 -2x^(1/2) +4x^(1/4) -4ln|1+x^(1/4)| + C'