# Convergence of integral

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1. Jan 30, 2015

### Neil21

1. The problem statement, all variables and given/known data
Find whether the integral is convergent or not, and evaluate if convergent.

2. Relevant equations
integral 1/sqrt(x^4+x^2+1) from 1 to infinity

3. The attempt at a solution
1/sqrt(x^4+x^2+1)<1/sqrt(x^4)
1/sqrt(x^4)=1/x^2 which is convergent for 1 to infinity and is 1
therefore, the original function is convergent
I can't seem to find the antiderivative of the original function after this.

2. Jan 30, 2015

### MostlyHarmless

Try completing the square

3. Jan 30, 2015

### Neil21

Ok, so I complete the square and got $$\int \frac{1}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}}$$ . Im still not seeing a way find an antiderivative though.

4. Jan 30, 2015

### Staff: Mentor

Don't forget the dx.
$$\int \frac{dx}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}}$$

I haven't worked this through, yet, but here's what I would do, FWIW.
Let u = x2 + 1/2
So $x = \sqrt{u - 1/2}$
Then du = 2xdx
So $dx = \frac{du}{2x} = \frac{du}{2\sqrt{u - 1/2}}$

After making the substitution and getting everything in terms of u and du, I would try for a trig substitution. I think that might work.

5. Jan 30, 2015

### Staff: Mentor

You don't need to integrate the exact expression. If the 3/4 wasn't in there, would the integral be convergent?

Chet

Last edited: Jan 30, 2015
6. Jan 30, 2015

### MostlyHarmless

Trig substitution is always my first instinct when I see a fraction with higher powers of x.

7. Jan 31, 2015

### Staff: Mentor

@chet, I think he already established that in the opening post.

8. Jan 31, 2015

### MostlyHarmless

What I first saw was rewriting the expression as $${\frac{1}{(x^4+x^2+1)^{1/2}}}={\frac{1}{((x^2+({\frac{1}{2}))^2}+(\frac{\sqrt{3}}{2})^2)^{1/2}}}$$.
Which I'm pretty sure can be solved with a trig sub.

9. Jan 31, 2015

### Staff: Mentor

Ooops. Missed that. Sorry guys.

Chet

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