Integral $\int_0^{\infty} \frac{x^{a-1}}{1+x} dx $ converges? For what values ​​of $a \in \mathbb{R}$ the following integral converges?
$$\int_0^{\infty} \frac{x^{a-1}}{1+x}\ dx $$
I tried to compute the integral but I stuck solving and then I tried to compare in various regions but I stuck, some help for this please.

Roughly speaking, if $a\le 0$, then there is trouble at $0$. If $a\ge 1$, there is trouble "at" infinity. And in the interval $(0,1)$ there is no trouble anywhere.
Case 1: Let $a\le 0$. Then for $0\lt x\le 1$ we have $\frac{x^{a-1}}{1+x} \ge \frac{1}{2x}$.
But $\int_0^1 \frac{1}{2x}\,dx$ diverges, and therefore by Comparison so does our integral. 
Case 2: Let $a\ge 1$. Then for $x\ge 1$ we have $\frac{x^{a-1}}{1+x} \ge \frac{1}{1+x}\ge \frac{1}{2x}$. But $\int_1^\infty \frac{1}{2x}\,dx$ diverges, and therefore by Comparison so does our integral. 

Case 3: Let $0\lt a\lt 1$. We show that both $\int_0^1 \frac{x^{a-1}}{1+x}\,dx$ and $\int_0^\infty \frac{x^{a-1}}{1+x}\,dx$ converge, and therefore so does our integral.
(i) Integral from $0$ to $1$: In the interval $0\lt x\le 1$, we have $0\lt \frac{x^{a-1}}{1+x}\lt \frac{1}{x^{1-a}}$. Since $1-a\lt 1$, the integral $\int_0^1 \frac{1}{x^{1-a}}\,dx$ converges, and therefore so does our integral. 
(ii) Integral from $1$ to $\infty$: In the interval $[1,\infty)$, we have $\frac{x^{a-1}}{1+x}\lt \frac{1}{x^{2-a}}$. But $2-a\gt 1$, so $\int_1^\infty \frac{1}{x^{2-a}}\,dx$ exists, and therefore by Comparison so does our integral. 
