I have no idea of Beta functions. Based on some properties of beta functions like, $B(m,n) = \int_{0}^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx \;\; ;B(m,n) = B(n,m)$
I arrived at $2B(m,n) = I+ \int_{1}^\infty \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}$ where I is my required integral. How to express the 2nd integral in terms of Beta function or is there any other way of solving. Need help. Thank You.
On
Let $I(m,n)$ be the integral defined by
$$I(m,n)=\int_0^1 \frac{t^{m-1}+t^{n-1}}{(1+t)^{n+m}}\,dt \tag 1$$
Enforcing the substitution $t\to \frac{1-t}{t}$ in $(1)$ reveals
$$\begin{align} I(m,n)&=\int_{1/2}^1 \left(t^{m-1}(1-t)^{n-1}+t^{n-1}(1-t)^{m-1}\right)\,dt \\\\ &=\int_{1/2}^1 t^{m-1}(1-t)^{n-1}\,dt +\int_{1/2}^1 t^{n-1}(1-t)^{m-1}\,dt \tag 2 \end{align}$$
Then, enforcing the substitution $t\to1-t$ in the second integral on the right-hand side of $(2)$ yields
$$\begin{align} I(m,n)&=\int_{1/2}^1 t^{m-1}(1-t)^{n-1}\,dt +\int_0^{1/2} t^{m-1}(1-t)^{n-1}\,dt\\\\ &=\int_0^1 t^{m-1}(1-t)^{n-1}\,dt\\\\ &=B(m,n) \end{align}$$
as was to be shown!
It is known that $ \displaystyle {\rm B}(x, y)=\int_{0}^{\infty}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\,dt $. From that we can obtain another great formula for the Beta Function which is $$ {\rm B}(x, y)=\int_{0}^{1}\frac{t^{x-1}+t^{y-1}}{\left ( 1+t \right )^{x+y}}\,dt $$
The proof is easy and it goes along these lines.
\begin{align*} {\rm B}(x, y) &= \int_{0}^{\infty} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, dt \\ &= \int_{0}^{1} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, dt + \int_{1}^{\infty} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, dt\\ &\overset{u=1/t}{=\! =\! =\! } \int_{0}^{1} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, dt + \int_{0}^{1} \frac{\left ( \frac{1}{t} \right )^{x-1}}{\left ( 1+\frac{1}{t} \right )^{x+y}} \frac{1}{t^2} \, dt\\ &= \int_{0}^{1} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, dt + \int_{0}^{1} \frac{t^{y-1}}{\left ( 1+t \right )^{x+y}} \, dt\\ &= \int_{0}^{1}\frac{t^{x-1}+t^{y-1}}{\left ( 1+t \right )^{x+y}}\,dt \end{align*}