Science:Math Exam Resources/Courses/MATH101 C/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 $\int _{1}^{\infty }{\frac {1}{x}}\;\mathrm {d} x$ 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 $x\in [1,\infty ),$ 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 $\int _{1}^{\infty }{\frac {K}{x}}\;\mathrm {d} x$ 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 $x\geq 1$, we have ${\frac {2}{x^{2}}}\leq {\frac {2}{1}}<2x,$ so we find

$${\frac {1}{x+{\frac {1}{x^{2}}}}}\geq {\frac {1}{x+2x}}.$$

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,

$$\int _{1}^{\infty }{\frac {1}{3x}}\;\mathrm {d} x$$

diverges, so does the integral

$$\int _{1}^{\infty }{\frac {x^{2}+1}{x^{3}+2}}\;\mathrm {d} x$$