Divergence And Convergence Integral Question: $e\int\limits^{\infty}_{0} \frac1{\sqrt{x}\left(1+x\right)}\,\mathrm{d}x$

Question

$\displaystyle\int\limits^{\infty}_{0} \dfrac{1}{\sqrt{x}\left(1+x\right)}\,\mathrm{d}x$

the question is

$\quad\text{Determine whether the integral is convergent or divergent.}$

How to approach this question?

Answer 1

For $x>1$, notice that

$$\frac1{\sqrt x(1+x)}<\frac1{\sqrt x(0+x)}=\frac1{x^{3/2}}$$

And for $0<x<1$, notice that

$$\frac1{\sqrt x(1+x)}<\frac1{\sqrt x(1+0)}=\frac1{\sqrt x}$$

Thus, we may conclude the integral converges, since

$$\int_0^\infty\frac1{\sqrt x(1+x)}\ dx<\int_0^1\frac1{\sqrt x}\ dx+\int_1^\infty\frac1{x^{3/2}}\ dx=2+2<\infty$$

Answer 2

By setting $x=z^2$, $$ \int_{0}^{+\infty}\frac{dx}{\sqrt{x}(1+x)} = \int_{0}^{+\infty}\frac{2\,dz}{1+z^2}=\color{red}{\pi} $$ is very much convergent.