Convergence of integral

[Neil21]

Homework Statement

Find whether the integral is convergent or not, and evaluate if convergent.

Homework Equations

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

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.

 

[MostlyHarmless]

Try completing the square

 

[Neil21]

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

 

[Mark44]

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

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.

 

[Chestermiller]

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

Chet

 

[MostlyHarmless]

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

 

[Mark44]

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

Chet

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

 

[MostlyHarmless]

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.

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.

 

[Chestermiller]

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

Ooops. Missed that. Sorry guys.

Chet