Prove convergence of $\int_1^\infty \frac 1 {x(\sqrt x + 1)} dx$

Question

Prove the convergence of $\int_1^\infty \frac 1 {x(\sqrt x + 1)} dx$

This was a question on an exam. I needed to prove that the above integral converges using the comparison test. I thought about using something along the lines of the integral of $1/(x^2+1)$ but I wasn't sure. I hope to learn from this so I can succeed on the final. Any help is appreciated.

Answer 1

Note that $\sqrt{x}+1 \ge \sqrt{x}$ implies that $\frac{1}{\sqrt{x}+1}\le\frac{1}{\sqrt{x}}$.

Thus, we have for $x>0$

$$\left|\frac{1}{x(\sqrt{x}+1)}\right|\le\frac{1}{x^{3/2}}$$

Since $$\int_1^{\infty}x^{-3/2}dx=2$$

then by the comparison test, we conclude that

$$\int_1^{\infty}\frac{dx}{x(x^{1/2}+1)}\,\,\text{converges}$$

Answer 2

Your integral is convergent since $\frac{1}{x(\sqrt{x}+1)}$ is equivalent to $\frac{1}{x^{3/2}}$ when $x$ is large and $3/2>1$ so by the Riemann criterion we have $\int_{1}^{+\infty} \frac{1}{x^{3/2}}$ is convergent so we get the convergence of the integral $\int_{1}^{+\infty}\frac{1}{x(\sqrt{x}+1)}.$