I have this improper integral and I want to know if it is convergent or divergent. My guess is that I should use the comparison theorem, because I have tried to integrate it and I found it difficult.
My attempt:
$\int_{1}^\infty \frac{x}{\sqrt{x^4+1}} dx$
So I start by choosing the "dominating" part on the bottom and keeping the top as it is.
$\int_{1}^\infty \frac{x}{\sqrt{x^4}} dx$
I can now remove the square root on the bottom.
$\int_{1}^\infty \frac{x}{x^2} dx$
Cancel one x.
$\int_{1}^\infty \frac{1}{x} dx$
Integrate.
$\ln{\lvert x\rvert}$ from 1 to $\infty$
$\ln{\lvert \infty \rvert} - \ln{\lvert 1 \rvert}$
This is divergent because $\ln{\lvert \infty \rvert}$ is $\infty$.
$\ln{\lvert \infty \rvert}$
Now we need to make sure that our original improper integral is greater than or equal to the simplified integral.
$\int_{1}^\infty \frac{x}{\sqrt{x^4+1}} dx \geq \int_{1}^\infty \frac{1}{x} dx \geq 0$ ?
The integral boundaries are the same so we only care about the functions.
$ \frac{x}{\sqrt{x^4+1}} \geq \frac{1}{x} \geq 0$ ?
Since everything is positive, we can cross-multiply.
$ x^2 \geq \sqrt{x^4+1} \geq 0$ ?
$ x^2$ is not greater than or equal to $\sqrt{x^4+1}$
I'm not sure how I should continue from here. I would really appreciate some help.
Yes, you need to find a function $f$ such that $\sqrt{x^4+1}\le f$, as you said, $f$ cannot be $x^2$, but what about $f=\sqrt{x^4+x^4}$ ?
Since $1\le x$, we have $\Rightarrow 1\le x^4\Rightarrow x^4+1\le x^4+x^4=2x^4$, therefore
$$\sqrt{x^4+1}\le \sqrt{2x^4}=\sqrt2~x^2$$
Now we can use comparison test:
$$\int_{1}^\infty \frac{x}{\sqrt{x^4+1}} dx\ge \int_{1}^\infty \frac{x}{\sqrt2~x^2} dx=\frac{1}{\sqrt2}\int_{1}^\infty \frac{1}{x} dx\to\infty$$
So the integral is divergent.