<s>[INST] How to show that $\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}$? From numerical evidence it appears that whenever the integral converges, $$J_a :=\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}.$$
For $a \in \mathbb{N}$, I was able to prove this using induction (see below). How can we prove it for non-integer $a$?

Integrating by parts,
  $$\begin{align}
J_n &= \left.-\frac{1}{n+1} \frac{1+x^n}{(1+x)^{n+1}}\right\lvert_0^1+ \frac{n}{n+1} \int_0^1 dx \frac{x^{n-1}}{(1+x)^{n+1}}
\\&=\frac{1}{n+1}\left(1-2^{-n} \right) + \frac{n}{n+1} \int_0^1 dx \left[ \frac{1+x^{n-1}}{(1+x)^{n+1}} - \frac{1}{(1+x)^{n+1}} \right]
\\&=\frac{n}{n+1}J_{n-1} + \frac{1}{n+1} \left[\left(1-2^{-n} \right) -\int_0^1 dx \frac{n}{(1+x)^{n+1}} \right]
\\&=\frac{n}{n+1}J_{n-1}.
\end{align}.
$$
  Because $J_0 = 1$, we therefore have $$J_n = \frac{1}{n+1}  \mathrm{for\,} n \in \mathbb{N}.$$

 [/INST] $$\begin{align*}
\int_{0}^{1}\frac{1+x^{a}}{\left(1+x\right)^{a+2}}dx
&=\int_{0}^{1}\frac{1}{\left(1+x\right)^{a+2}}dx+\int_{0}^{1}\frac{x^{a}}{\left(1+x\right)^{a+2}}dx \\
&=\left.-\frac{1+x}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1}+\left.\frac{\left(1+x\right)x^{a+1}}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1} \\
&=\frac{1}{a+1}.
\end{align*}$$
</s>