Question

求不定积分∫（（1-√（x+1））/（1+（x+1）^（1/3）））dx

Answer 1

这题要令(1+x)^(1/6)=t, 1+x=t^6. dx=6t^5dt
被积函数为：6t^5(1-t^3)/(1+t^2)=-6t^6+6t^4+6t^3-6t^2-6t+6+6t/(t^2+1)-6/(1+t^2)
积分=-6t^7/7+6t^5/5+6t^4/4-6t^3/3-6t^2/2+6t+3ln(t^2+1)-6arctant+C
=-6t^7/7+6t^5/5+3t^4/2-2t^3-3t^2+6t+3ln(t^2+1)-6arctant+C
最后把t=(1+x)^(1/6)代入

Answer 2

用换元法
令（x+1）^（1/6）=t
√（x+1）=t^3
x=t^6-1
dx=6t^5dt
∫（（1-√（x+1））/（1+（x+1）^（1/3）））dx
=∫(1-t^3)/(1+t^2)*6t^5dt
=6∫(t^5-t^8)/(1+t^2)dt
=6∫t^5/(1+t^2)dt-6∫t^8/(1+t^2)dt
=3∫t^4/(1+t^2)dt^2-6∫(t^8-1+1)/(1+t^2)dt
=3∫(t^4-1+1)/(1+t^2)dt^2-6∫(t^8-1+1)/(1+t^2)dt
=3∫[t^2-1+1/(1+t^2)]dt^2-6∫[(t^2-1)(t^4+1)+1/(1+t^2)]dt
=3/2t^4-3t^2+2ln(1+t^2)-6∫[t^6-t^4+t^2+1+1/(1+t^2)]dt
=3/2t^4-3t^2+2ln(1+t^2)-6/7*t^7-6/5t^5+2t^3+1+6arctant+C
自己反代吧

Answer 3

设t^6=x+1,x=t^6-1,dx=6t^5dt
原式=∫(1-t³)/(1+t²)*6t^5dt
=6∫(-t^8-t^5)/(1+t²)dt=1/6∫(-t^6+t^4-t³+t+1)-(1+t)/(1+t²)dt
=6{[1/7*t^7+1/5*t^5-1/4^t^4+1/2t^2+t]-arctant-1/2ln(1+t²)}+C
t=(1+x)^(1/6)代入
=6{[1/7*(1+x)^(7/6)+1/5*(1+x)^(5/6)-1/4^(1+x)^(2/3)+1/2(1+x)^(1/3)+(1+x)^(1/6)]-arctan(1+x)^(1/6)-1/2ln(1+(1+x)^(1/3))}+C