Integration of a func raised to a fraction

1. Aug 25, 2009

mojomike

1. The problem statement, all variables and given/known data
I need to solve the following integral:

{ x * ((1-x)/3)^(1/7) dx

where { represents the integral symbol b/c I don't know how to type that.

3. The attempt at a solution

I've tried integration by parts and don't get anywhere. None of the integration techiques I know seem to work and nothing I've found online says what to do when you have this kind of a situation.

Any help would be greatly appreciated. I'm lost. Thanks everyone!!

,Mike
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 25, 2009

LeonhardEuler

Can you think of a substitution that would make the (1/7) power part easy to integrate?

3. Aug 25, 2009

mg0stisha

I don't think he's coming back (as that was his only post), but I'm relatively new to calc and would like help solving this. Anyone willing to help?

4. Aug 25, 2009

mojomike

Thanks for the response,

Well, I tried to substitute that inside the parenthesis, but there's that blasted extra x outside of it to deal with. In integrating by parts, I made x the integration term(v) and the other part the differential term, but the result is just another integral only more complicated.

5. Aug 25, 2009

VietDao29

No, you don't use Integration by Parts in this problem.

To type maths beautifully, one should learn how to https://www.physicsforums.com/showthread.php?t=8997". And you can begin to dig in the awesome world of LaTeX today.

$$\int x \left( \frac{1 - x}{3} \right) ^ {\frac{1}{7}} dx$$, right?

The first step is to take all constants out of it:

$$\frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx$$

The product of (1 - x)1 / 7 and x is a little bit complicated to integrate, right? Because you just cannot expand the part (1 - x)1 / 7.

If the problem was: $$\int x \left( 1 - x \right) ^ {2} dx$$, then it would be much easier, because we can expand (1 - x)2 = 1 - 2x + x2. And the whole integral would become: $$\int (x ^ 3 - 2 x ^ 2 + x) dx$$, which should be like, a piece of cake.

So, back to the problem, we then think about a simple u-substitution, which eliminates the needs to expand (1 - x)1 / 7. Let's see if you can figure out what u should be. :)

Last edited by a moderator: Apr 24, 2017
6. Aug 25, 2009

mg0stisha

Would it be u=1/7?

7. Aug 25, 2009

Elucidus

Letting u equal a constant implies du = 0 and the integral is trivial. The difficulty with the integrand was the (1 - x) being raised to a fractional power. If the base were simpler....

--Elucidus

8. Aug 26, 2009

mg0stisha

So right now we have $$\frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx$$ correct?
i guess we could distribute the x in front of the parentheses so we'll have$$\frac{1}{\sqrt[7]{3}} \int \left( x - x^2 \right) ^ {\frac{1}{7}} dx$$? then simplify that and pull the negative sign out front? but then there's no substitution...

the only other idea would be to make a substitution so that u=(1-x)?

9. Aug 26, 2009

darkmagic

Bring all constants outside then integrate. Use integration by parts. u=x, du=dx,
dv=(1-x)^1/7 dx, v=-7/8(1-x)^8/7. Any corrections if I integrated it wrong.

10. Aug 26, 2009

Cyosis

This is not correct, you cannot distribute unequal powers. Note that $x=(x^7)^{\frac{1}{7}}$. Now that they are equal powers you can distribute, which would yield $x(1-x)^{\frac{1}{7}}=(x^7-x^8)^{\frac{1}{7}}$. This doesn't make things any easier.
As hinted before use $u=1-x$.

11. Aug 26, 2009

Elucidus

To go from $dv=(1-x)^{1/7}dx \text{ to }v=-\frac{7}{8}(1-x)^{8/7}$ requires integration by substitution (in fact, using the same substitution that would solve it directly). The integral can be done more simply than by integration by parts - although "by parts" is valid.

--Elucidus

12. Aug 26, 2009

mg0stisha

I just can't figure it out. I've never formally taken a calculus class, just taught myself out of a cheap self teach book from barnes and noble. I substituted so that u = (1-x) and now have $$\frac{1}{\sqrt[7]{3}} \int x \left( u\right) ^ {\frac{1}{7}} dx$$. And, once again i'm stuck.

13. Aug 26, 2009

LeonhardEuler

If u= 1- x, then what is 'x'? You need to make the substitution everywhere there is an 'x'. This means you must also integrate du instead of dx.

14. Aug 26, 2009

Cyosis

As pointed out you need to express every x in terms of u, including the differential. If you substitute u=f(x), then du=f'(x)dx. In your case f(x)=1-x. Can you continue from here?

15. Aug 26, 2009

mg0stisha

u= 1-x
du= -1
correct?

16. Aug 26, 2009

Cyosis

No, if f(x)=1-x and du=f'(x)dx, then f'(x)=? and du=?

17. Aug 26, 2009

mg0stisha

The derivative of 1-x isnt -1?

18. Aug 26, 2009

Cyosis

Yes that part is correct, however incomplete. You want to express the differential, du in terms of the differential dx.

19. Aug 26, 2009

mg0stisha

du/dx = -1? i have no clue why i'm so lost right now.

20. Aug 26, 2009

Cyosis

It's not much more than plug and play. Two differentials are related as $du=\frac{du}{dx}dx$. This is the exact same as the one in my qu0te, just with a different notation. Use these properties to find du in terms of x.