Does the integral converge? $\int_1^\infty \frac{1}{x(x^2+1)}dx$

Question

Does the integral converge? $$\int_1^\infty \frac{1}{x(x^2+1)}dx$$

Well, I did show that it converge by finding the indefinite integral first, and getting to $\lim =\frac{\ln 2}{2}$. Which means it converges.

But I thought about another approach, a faster one, I would love to get your opinion.

I thought about using D'Alembert's rule, with the integral $\int \frac{1}{x}$, and saying that they are positive for every $x \in [1,\infty]$, and because the limit now will be finite, I can say that because $\int \frac{1}{x}$ diverges then $\int_1^\infty \frac{1}{x(x^2+1)}dx$ too.

Answer 1

Since $x\geq 1$ in the domain of integration, we know that $$\frac{1}{x(x^2+1)}\leq\frac{1}{x^2+1}.$$ It is well known that $$\int_1^\infty \frac{1}{1+x^2}\,dx\leq\int_{-\infty}^\infty\frac{1}{1+x^2}\,dx=\pi<\infty.$$ So the integral converges. You cannot say that since $\int\frac{1}{x}\,dx$ diverges, $\int_1^\infty \frac{1}{x(1+x^2)}\,dx$ does too since $1/x\geq\frac{1}{x(x^2+1)}$ on the domain of integration.

Answer 2

Hint : $\dfrac{1}{x(x^2+1)} < \dfrac{1}{x^2+1}$.