# Math Help - Improper Integral

1. ## Improper Integral

$\int_1^\infty \frac{1}{(x^4+x+1)^{\frac{1}{3}}}dx$

I don't think that I can evaluate the integral easily, so I'm looking for a comparative function. If I say that:

$x^4 + x + 1 \geq x for [1,\infty)$

then I can also say that

$\frac{1}{x^4 + x + 1} \leq \frac{1}{x}$

$\frac{1}{(x^4+x+1)^{\frac{1}{3}}} \leq \frac{1}{x^{\frac{1}{3}}}$

The problem is that
$\int_1^\infty \frac{1}{x^{\frac{1}{3}}} = \infty$
which doesn't help me in determining the value of the original integral.

Can anybody see what I am doing wrong or recommend another function to use as a comparison?

Thanks.

2. ## Re: Improper Integral

How about comparing 1/(u^1/4) with 1/(u^1/3)

(where u = x^4 + x + 1

3. ## Re: Improper Integral

Thanks, but I don't see how this solves the problem. Changing the exponent from 1/3 to 1/4 still leaves me with an integral that I can't evaluate easily, and doesn't give me a function that I can show goes to infinity or to a finite number. What am I missing?

Thanks again!

4. ## Re: Improper Integral

Originally Posted by joatmon
$\int_1^\infty \frac{1}{(x^4+x+1)^{\frac{1}{3}}}dx$

I don't think that I can evaluate the integral easily, so I'm looking for a comparative function. If I say that:

$x^4 + x + 1 \geq x for [1,\infty)$

then I can also say that

$\frac{1}{x^4 + x + 1} \leq \frac{1}{x}$

$\frac{1}{(x^4+x+1)^{\frac{1}{3}}} \leq \frac{1}{x^{\frac{1}{3}}}$

The problem is that
$\int_1^\infty \frac{1}{x^{\frac{1}{3}}} = \infty$
which doesn't help me in determining the value of the original integral.

Can anybody see what I am doing wrong or recommend another function to use as a comparison?

Thanks.
For x>1 is $x^{4}+x+1 > x^{4}$ so that is also $\frac{1}{x^{4}+x+1}<\frac{1}{x^{4}}$ and the integral $\int_{1}^{\infty} \frac{dx}{x^{\frac{4}{3}}}=3$ converges , so that...

Kind regards

$\chi$ $\sigma$

Thank you!