Question

求助一道高数题，会的帮下忙〜一定有好评！

Answer 1

∫dx/[x^(1/2)+x^(1/3)]，u=x^1/6，x=u^6，dx=6u^5du
= 6∫u^5du/(u³+u²)
= 6∫u³du/(u+1)
= 6∫[u²-u-1/(u+1)+1]du
= 6[u³-u²-ln|u+1|+u] + c
= 2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|1+x^(1/6)| + c

追答

这是不定积分的求法

定积分就只要在
= 6[u³-u²-ln|u+1|+u] + c
这一步把u回代再带入数据就行了

哦，对了，要去掉那个C

追问

谢谢！

追答

不客气

Answer 2

令t=x^1/6,x=t^6,dx=6t^5
原式=S(0,1)6t^5dt/(t^3+t^2)=S(0,1)6t^3dt/(t+1)=6S(0,1)(t^3+t^2-t^2-t+t+1-1)dt/(t+1)=6{1/3-1/2+1-ln2}=5-6ln2

Answer 3