Convergence test for an integral of bounds $0$ to $1$

Question

How to prove that the integral: $\int_{0}^{1} \frac{dx}{2\sqrt{x}(x+1)}$ converges using the convergence test?

I know that $\int_{0}^{1} \frac{1}{x^{\alpha}} d x$ converges $\iff \alpha < 1$. But in my case, the denominator does not look like that, and I find it impossible to transform it into this form.

Is there another test to prove the convergence of this integral?

Answer 1

$0 <\frac 1 {2 \sqrt x (x+1)} <\frac 1{2\sqrt x}$ and $\int_0^{1} \frac 1{2\sqrt x}dx$ is convergent. Hence the given integral is convergent.

Answer 2

Besides Kavi's answer, observe that

$$\int\frac{dx}{2\sqrt x(x+1)}=\arctan\sqrt x+C\implies\int_0^1\frac{dx}{2\sqrt x(x+1)}=$$

$$=\left.\arctan\sqrt x\right|_0^1=\arctan1-\arctan0=\frac\pi4$$

Answer 3

$\frac{1}{2\sqrt{x}(x+1)} \sim \frac{1}{2\sqrt{x}}$ when $x \rightarrow 0$