Question

What is the integral of `int x^(1/3)/[x^(1/3)+x^(1/4)]` ?

Answer 1

`12{y^12/12-y^11/11+y^10/10-y^9/9+...+y^2/2-y+ln|(y-1)|}+c`,

where, `y=x^(1/12)`.

Explanation:

Let, `I=intx^(1/3)/(x^(1/3)+x^(1/4))dx`.

To remove the radicals, subst. `x=y^12. :. dx=12y^11dy`.

`:. I=inty^4/(y^4+y^3)*12y^11dy`,

`:. I=12inty^12/(y+1)dy............(ast)`.

Now, consider the sum of first `12` terms of the following

geometric series having the `1^(st)` term `1` and common ratio `(-r) :`

`:. 1-r+r^2-r^3+...-r^11=[1*{(-r)^12-1}]/{(-r)-1}, i.e.,`

`(r^12-1)/(r+1)=r^11-r^10+r^9-r^8+...+r-1`.

Utilising this, we find,

`y^12/(y+1)={(y^12-1)+1}/(y+1)=(y^12-1)/(y+1)+1/(y+1)`,

`=y^11-y^10+y^9-y^8+...+y-1+1/(y-1)`.

`:. I=12int{y^11-y^10+y^9-y^8+...+y-1+1/(y-1)}dy`,

`=12{y^12/12-y^11/11+y^10/10-y^9/9+...+y^2/2-y+ln|(y-1)|}+c`,

where, `y=x^(1/12)`.

Answer 2

I have tried ,but tired....!
`I=12[x/12-x^(11/12)/11+x^(10/12)/10-x^(9/12)/9+x^(8/12)/8-x^(7/12)/7+x^(6/12)/6-x^(5/12)/5+x^(4/12)/4-x^(3/12)/3+x^(2/12)/2-x^(1/12)]-12ln|x^(1/12)+1|+C`

Explanation:

Here,

`I=intx^(1/3)/(x^(1/3)+x^(1/4))dx`

Let,

`x=u^12=>dx=12u^11du and u=x^(1/12)`

`=>x^(1/3)=(u^12)^(1/3)=u^4andx^(1/4)=(u^12)^(1/4)=u^3`

So,

`I=intu^4/(u^4+u^3)xx12u^11 du`

`=12intu^15/(u^3(u+1))du`

`=12intu^12/(u+1)du`

`=12int [(u^12+1)/(u+1)-1/(u+1)]du`

`I=12int(u^12+1)/(u+1)du-12ln|u+1|`

`I=12I_1-12ln|u+1|...to(A)`

`Now,(u^12 +1)`=#(u+1)(u^11-u^10+u^9-u^8+u^7-u^6...+u^3- u^2+u-1)#

So,`I_1`

`=int[u^11-u^10+u^9-u^8...+u^5-u^4+u^3-u^2+u-1]du`

#I_1=u^12/12-u^11/11+u^10/10-u^9/9+u^8/8-u^7/7+u^6/6- u^5/5+u^4/4-u^3/3+u^2/2-u+c#

Hence, from `(A)`

#I=12[u^12/12-u^11/11+u^10/10-u^9/9+u^8/8-u^7/7+u^6/6- u^5/5+u^4/4-u^3/3+u^2/2-u]-12ln|u+1|+C#

Subst.back, `u=x^(1/12)`

#I=12[x/12-x^(11/12)/11+x^(10/12)/10-x^(9/12)/9+x^(8/12)/8- x^(7/12)/7+x^(6/12)/6-x^(5/12)/5+x^(4/12)/4- x^(3/12)/3+x^(2/12)/2-x^(1/12)]-12ln|x^(1/12)+1|+C#