# 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.