# Math Help - Comparison of Improper Integrals

1. ## Comparison of Improper Integrals

Problem number 1: $\int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$

and problem number 2: $\int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$

2. Originally Posted by theskyidfalling
Problem number 1: $\int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$

and problem number 2: $\int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
Problem number 1: $\int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx < \int_1^{\infty} \frac{1}{x} dx$(**)

Problem number 2: $\int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx < \int_1^{\infty} \frac{25}{5x^3-2x^2} dx$

Both the integrals I gave are integrable and converge. (No they don't - not **)

3. Thanks! But how does the first one converge? I thought it diverged since $\int_1^{\infty} \frac{1{}}{x} dx$ goes to infinity? And how do you come up with the second integral?

4. Originally Posted by theskyidfalling
Thanks! But how does the first one converge? I thought it diverged since $\int_1^{\infty} \frac{1{}}{x} dx$ goes to infinity? And how do you come up with the second integral?
Opps, it doesn't. (Do over)

$\int_1^{\infty} \frac{1}{x+2} dx < \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$

5. On both, look at the long term behaviour of each. For the first

$\frac{x^2+1}{x^3+3x+2} \approx \frac{1}{x}$ and the second $\frac{5x+2}{x^4+8x^2+4} \approx \frac{5}{x^3}$.

So we need inequalities that are integrable. For the first

$\frac{1}{x} \le\frac{x^2+1}{x^3+3x+2}$ isn't true since

$x^3+3x+2 \le x (x^2+1)$ isn't true because $3x+2 \le x$ isn't true, so we need little more to have something true for all $x \ge1$. Try

$\frac{1}{x+a} \le \frac{x^2+1}{x^3+3x+2}$

so

$x^3+3x+2 \le (x+a) (x^2+1)$ so $3x+2 \le ax^2+x+a$ or $0 \le ax^2-2 x+a-2$. We see that if $a = 2$ then $0 \le 2x^2-2 x = 2x(x-1)$ which is true.

Same for the second problem.

$\frac{5x+2}{x^4+8x^2+4} \le \frac{5}{x^3}$ isn't true so we try

$\frac{5x+2}{x^4+8x^2+4} \le \frac{5}{x^3+ax^2}$ and find an a where this is true.

6. Originally Posted by theskyidfalling
Hi so I was just wondering if anyone could help me figure out how to solve these two particular problems. Thank you!

Problem number 1: $\int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$
$u= x^3+ 3x+ 2$ look like an obvious substitution!

and problem number 2: $\int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
$x^4+ 8x^2+ 4= x^4+ 8x^2+ 16- 12= (x^2+ 4)^2- 12$ which factors as $(x^2+4- 2\sqrt{3})(x^2+ 4+ 2\sqrt{3})$. See if that helps.

7. Originally Posted by theskyidfalling

Problem number 1: $\int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$
$\frac{x^{2}+1}{x^{3}+3x+2}>\frac{x^{2}}{x^{3}+3x^{ 3}+2x^{3}}>\frac{1}{6x},$ thus the integral diverges.

Originally Posted by theskyidfalling

and problem number 2: $\int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
$\frac{5x+2}{x^{4}+8x^{2}+4}\le\frac{5x+2x}{x^{4}}= \frac{7}{x^{3}},$ thus the integral converges.