# Homework Help: Weird Integral

1. Feb 3, 2010

### Rats_N_Cats

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

An Integral : $$\int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x}$$

2. Relevant equations

The Standard integrals.

3. The attempt at a solution

I'm aware that integrals like this become very easy after a clever substitution...but maybe I'm not that clever so I can't even start it. If anyone shows me the first step I'll try to take it from there.

Last edited: Feb 3, 2010
2. Feb 3, 2010

### ideasrule

$$\int \frac{1}{x^n(1+x^n)^{1/n}} \mathrm{d}x$$

Last edited: Feb 3, 2010
3. Feb 3, 2010

### Rats_N_Cats

You've left the dx at the bottom
but how will it become $(1+x^n)^{n}$??? Bringing it to the top will change the sign of the exponent, right?

4. Feb 3, 2010

### ideasrule

Yes, I got confused. Sorry about that.

5. Feb 3, 2010

### Rats_N_Cats

I tried letting xn = t, but that ended up with $$\frac{1}{n} \int \frac{1}{\sqrt[n]{t^2+t}}\:\mathrm{d}t$$, And I don't see how to do it.
Then I tried letting xn+1 = tn, and got something similarly unsolvable. Can anyone tell me what's the right substitution in this case?

Last edited: Feb 3, 2010
6. Feb 4, 2010

### Char. Limit

Here's something:

$$\int x^{-n}(1+x^n)^{-1/n} \mathrm{d}x$$

And integrate by parts from there. Dunno if it works, though. I tried it on W-A and they used a weird substitution within a substitution.

7. Feb 4, 2010

### Rats_N_Cats

I did that, letting (1+xn)-1/n as first function, and I ended up with :
$$\frac{x^{1-n}}{1-n}\,(1+x^n)^{-1/n} + \frac{x}{1-n} + \frac{x^{n+1}}{1-n^2} + C$$
Is that correct?

8. Feb 4, 2010

### Char. Limit

Maybe. I'm too tired to check now. It looks right.

9. Feb 5, 2010

### Rats_N_Cats

Will anyone please confirm if my answer is correct or not? This problem's been bugging me for quite some time.

10. Feb 5, 2010

### nobahar

11. Feb 5, 2010

### Zayer

try trigonometric substitution

12. Feb 6, 2010

### nobahar

Hello!
What trig substitution can you make? PM me if the OP wants to work it out themselves.
Thanks!