Definite integral with parameter

For the values of $a$ for which the integral converges, evaluate the integral:

$$\mathcal{J}=\int_0^1 \frac{1+x^a}{(1+x)^{a+2}}\, {\rm d}x$$

Solution

Lemma: It is pretty known that a form of the Beta function is:

$${\rm B}(x, y)=\int_{0}^{\infty}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\,{\rm d}t$$

Hence:
\begin{align*}
B(x, y)=\int_{0}^{\infty}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t &=\int_{0}^{1}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t + \int_{1}^{\infty}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t \\
 &\overset{u=1/t}{=\! =\! =\!}\int_{0}^{1}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t + \int_{0}^{1}\frac{\left ( \frac{1}{u} \right )^{x-1}}{\left ( 1+ \frac{1}{u} \right )^{x+y}}\cdot \frac{1}{u^2}\, {\rm d}u\\
 &= \int_{0}^{1}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t + \int_{0}^{1}\frac{t^{y-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t\\
 &=\int_{0}^{1}\frac{t^{x-1}+t^{y-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t
\end{align*}

 Our integral obeys the above formula hence:

$$ \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx={\rm B}\left ( 1, a+1 \right )=\frac{\Gamma (1)\Gamma (a+1)}{\Gamma (a+2)}=\frac{\cancel{\Gamma (a+1)}}{(a+1)\cancel{\Gamma (a+1)}}=\frac{1}{a+1}$$

Note: The integral converges when $a> -1$. If $a<-1$ then it divirges since it behaves like $1/x^a$. Obviously when $a=-1$ the integral also divirges.