Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 09/Solution 1

Following the hint, we have $$\frac{x^2+1}{x^3+2}=\frac{1}{x+\frac{2}{x^2}}+\frac{1}{x^3+2}.$$

Given that the first term in the sum above looks close to $\frac{1}{x}$, and we know that $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x}\mathrm{d}x$ diverges, we should try to prove that the integral above diverges as well. To do this, let us try to bound the integrand below by a function that diverges. Since $x\in [1,\mathrm{\infty }),$ we have

$$\frac{1}{x+\frac{2}{x^2}}+\frac{1}{x^3+2}>\frac{1}{x+\frac{2}{x^2}}.$$

If we could say next that $\frac{1}{x+\frac{2}{x^2}}>\frac{1}{x}$, we would be done. But, as we can see by substituting some values for $x$ this is not true! However, if we can show that $\frac{1}{x+\frac{2}{x^2}}>\frac{K}{x}$, for some $K>0$, we are still ok, since the integral $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{K}{x}\mathrm{d}x$ diverges as well. To achieve this, we should make sure that $\frac{2}{x^2}$ is not too large, so that the ratio $\frac{1}{x+\frac{2}{x^2}}$ is not much smaller than $\frac{1}{x}$. Since $x\ge 1$, we have $\frac{2}{x^2}\le \frac{2}{1}<2x,$ so we find $$\frac{1}{x+\frac{1}{x^2}}\ge \frac{1}{x+2x}.$$

Thus we have shown that the integrand $\frac{x^2+1}{x^3+2}$ is bounded below by $\frac{K}{x}$, with $K=\frac{1}{3}$, on the domain of integration. Since the integral of the lower bound, $$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{1}{3x}\mathrm{d}x$$ diverges, so does the integral $$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{x^2+1}{x^3+2}\mathrm{d}x$$