# Finding convergence/divergence of improper integral

Determine the convergence or divergence of this following improper integral:

$$\int_2^∞ \frac {1}{(x^3+7)^{\frac{1}{3}}}$$

So I'm trying to find something easy to compare this to, any help?

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Well, as I'm sure you can tell, it is a lot like 1/x. So, I would try to find something that is close to 1/x to which you can compare this.

Why can't I just compare it to 1/x?

I don't know if you have gone over this, but you can compare it to

$$\int\limits_1^\infty {\frac{1}{{{x^n}}}}$$

If n<2, it is divergent. If n≥2, then it is convergent.

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Well,
$$\frac{1}{(x^3+7)^{\frac{1}{3}}} < \frac{1}{x}$$

isn't it? Since [
$$\int_2^∞ \frac {1}{x}$$
diverges, this isn't any help.

So, I'd try showing that
$$(x^3+7)^{1/3} < x + b$$

where b is a number. (In particular, b is a number that appears in your problem).

Then realize that

$$\int_2^∞ \frac {1}{x+b}$$

diverges, and you're done.

How can I show that $$\int_2^∞ \frac {1}{x+7}$$ is divergent?

How can I show that $$\int_2^∞ \frac {1}{x+7}$$ is divergent?
First of all, I was kind of wrong when I said that "b" was in your problem. Use $$7^{1/3}$$ in stead of $$7$$. As for showing that

$$\int_2^∞ \frac {1}{x+7^{1/3}}$$

is divergent, is there some theorem in your book that would help? I don't know if there is or not. But, you could say that

$$lim \frac{1}{x+7^{1/3}} = \frac{1}{x}$$

and argue from there. Or, you could just integrate

$$\int_2^∞ \frac {1}{x+7^{1/3}}$$

You have done so to show that

$$\int_2^∞ \frac {1}{x}$$

is divergent, right? Just use a change of variables.