Problem number 1: $\displaystyle \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$
and problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
Problem number 1: $\displaystyle \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx < \int_1^{\infty} \frac{1}{x} dx$(**)
Problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx < \int_1^{\infty} \frac{25}{5x^3-2x^2} dx$
Both the integrals I gave are integrable and converge. (No they don't - not **)
On both, look at the long term behaviour of each. For the first
$\displaystyle \frac{x^2+1}{x^3+3x+2} \approx \frac{1}{x}$ and the second $\displaystyle \frac{5x+2}{x^4+8x^2+4} \approx \frac{5}{x^3}$.
So we need inequalities that are integrable. For the first
$\displaystyle \frac{1}{x} \le\frac{x^2+1}{x^3+3x+2} $ isn't true since
$\displaystyle x^3+3x+2 \le x (x^2+1)$ isn't true because $\displaystyle 3x+2 \le x $ isn't true, so we need little more to have something true for all $\displaystyle x \ge1 $. Try
$\displaystyle \frac{1}{x+a} \le \frac{x^2+1}{x^3+3x+2} $
so
$\displaystyle x^3+3x+2 \le (x+a) (x^2+1)$ so $\displaystyle 3x+2 \le ax^2+x+a$ or $\displaystyle 0 \le ax^2-2 x+a-2$. We see that if $\displaystyle a = 2$ then $\displaystyle 0 \le 2x^2-2 x = 2x(x-1)$ which is true.
Same for the second problem.
$\displaystyle \frac{5x+2}{x^4+8x^2+4} \le \frac{5}{x^3}$ isn't true so we try
$\displaystyle \frac{5x+2}{x^4+8x^2+4} \le \frac{5}{x^3+ax^2}$ and find an a where this is true.
$\displaystyle u= x^3+ 3x+ 2$ look like an obvious substitution!
$\displaystyle x^4+ 8x^2+ 4= x^4+ 8x^2+ 16- 12= (x^2+ 4)^2- 12$ which factors as $\displaystyle (x^2+4- 2\sqrt{3})(x^2+ 4+ 2\sqrt{3})$. See if that helps.and problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
$\displaystyle \frac{x^{2}+1}{x^{3}+3x+2}>\frac{x^{2}}{x^{3}+3x^{ 3}+2x^{3}}>\frac{1}{6x},$ thus the integral diverges.
$\displaystyle \frac{5x+2}{x^{4}+8x^{2}+4}\le\frac{5x+2x}{x^{4}}= \frac{7}{x^{3}},$ thus the integral converges.