# Comparison of Integrals

• Jacobpm64
In summary, the integrals in the box above converge for some values of p and diverge for other values of p.f

## Homework Statement

Use the box and the behavior of rational and exponential functions as $$x \rightarrow \infty$$ to predict whether the integrals converge or diverge.

Here is the box:
$$\int^\infty_1 \frac{1}{x^p} dx$$ converges for p > 1 and diverges for p < 1.

$$\int^1_0 \frac{1}{x^p} dx$$ converges for p < 1 and diverges for p > 1.

$$\int^\infty_0 e^{-ax} dx$$ converges for a > 0.

Problem 1:
$$\int^\infty_1 \frac{x^2}{x^4 + 1} dx$$

Problem 2:
$$\int^\infty_1 \frac{x^2 - 6x + 1}{x^2 + 4} dx$$

## Homework Equations

The ones in the box above.

## The Attempt at a Solution

Problem 1:
I know that this integral is less than $$\int^\infty_1 \frac{1}{x} dx$$. I also know that $$\int^\infty_1 \frac{1}{x} dx$$ diverges. This does not help me though because I can not use a diverging integral to say that a smaller integral is also diverging. This is where I'm confused.

Problem 2:
I know that this integral is less than $$\int^\infty_1 1 dx$$ I also know that $$\int^\infty_1 1 dx$$ diverges. This does not help me though because I can not use a diverging integral to say that a smaller integral is also diverging. So, I'm confused at the same place on this problem.

For problem 1, isn't (x^2)/((x^4)+1) always less than 1/(x^2) for values of x > 1?

For problem 2, what does the integrand approach as x approaches infinity?

$$\frac{x^2}{x^4+1}< \frac{x^2}{x^4}= \frac{1}{x^2}$$

For the second one, divide first:
[tex]\frac{x^2+ 6x+ 1}{x^2+ 4}= 1+ \frac{6x- 3}{x^2+ 4}[/itex]
What does that first "1" tell you?

Oops, I copied down the wrong first problem.

The second one was correct though.

I understand the first one, thanks for that.

I still don't understand what you mean by what does the first "1" tell me.

Wait, does that mean.. that my integral is bigger than the integral of 1, so if the integral of 1 diverges, I can say that my integral also diverges!

woot!

Thanks!

I'm going to post the correct problem for #1, in a new post. Thanks.