Improper Integral

Relevant wiki: Integration Tricks

$\begin{aligned} I & = \int_0^\infty \frac {dx}{(1+x)^2(1+x^\alpha)} & \small \color{#3D99F6} \text{Using the identity: }\int_0^\infty f(x)\ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{(1+x)^2(1+x^\alpha)} + \frac 1{x^2\left(1+\frac 1x\right)^2 \left(1+\frac 1{x^\alpha}\right)} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{(1+x)^2(1+x^\alpha)} + \frac {x^\alpha}{(1+x)^2(1+x^\alpha)} \right) dx \\ & = \frac 12 \int_0^\infty \frac {dx}{(1+x)^2} \\ & = \frac 12 \left[ - \frac 1{1+x} \right]_0^\infty = \frac 12 = \boxed{0.5} \end{aligned}$

$\begin{aligned} I&=\int_0^\infty\frac{dx}{(1+x)^2(1+x^\alpha)}\\ &=\int_0^1\frac{dx}{(1+x)^2(1+x^\alpha)}+\color{#3D99F6}\large\int_1^\infty\frac{dx}{(1+x)^2(1+x^\alpha)}&\color{#3D99F6}\text{Transform variables to }t=\frac{1}{x},\ dt=-\frac{dx}{x^2}\\ &=\int_0^1\frac{dx}{(1+x)^2(1+x^\alpha)}+\large\int_0^1\frac{t^\alpha}{(1+t)^2(1+t^\alpha)}dt\\ &=\int_0^1\frac{dx}{(1+x)^2}\\ &=\left[-\frac{1}{1+x}\right]_0^1=\frac12=\boxed{0.5} \end{aligned}$

All these solutions are pretty nice, but for a JEE shortcut, simply put alpha=1 and solve!!!!