Integral of $R(R^2+y^2)^{-3/2}$ with respect to $y$ $$\int_0^\infty \frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}dy$$
The indefinite integral seems to be $$\frac{-R}{\sqrt{R^2+y^2}}+C$$ 
$R$ is a constant
 A: We assume that $R$ is positive. 
Let $y=R\tan\theta$ or equivalently $\theta=\arctan(y/R)$.  As $y$ ranges from $0$ to $\infty$, the number $\theta$ ranges from $0$ to $\pi/2$.
Note that $\sqrt{R^2+y^2}\left(R^2+y^2\right)= R^3 \sec^3\theta$ and $dy=R\sec^2\theta\,d\theta$. We leave the rest to you.   
A: Using $y=Rx$, then $x=\tan(\theta)$ we get
$$
\begin{align}
\int_0^\infty\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y
&=\frac1R\int_0^\infty\frac{\mathrm{d}x}{\sqrt{1+x^2}\left(1+x^2\right)}\\
&=\frac1R\int_0^{\pi/2}\cos(\theta)\,\mathrm{d}\theta\\
&=\frac1R
\end{align}
$$
It seems that you have miscomputed the primitive.

Computing the Primitive
Combining the previous substitutions with $y=R\tan(\theta)$, we get
$$
\begin{align}
\int\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y
&=\frac1R\int\cos(\theta)\,\mathrm{d}\theta\\
&=\frac1R\sin(\theta)+C\\
&=\frac1R\frac{y}{\sqrt{R^2+y^2}}+C
\end{align}
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\overbrace{\color{#66f}{\large\int_{0}^{\infty}{R \over \pars{R^{2} + y^{2}}^{3/2}}\,\dd y}}
^{\ds{y = \verts{R}\sinh\pars{\theta}}}\ =\
{R \over \verts{R}^{2}}\int_{0}^{\infty}\sech^{2}\pars{\theta}\,\dd\theta
=\left.{1 \over R}\,\tanh\pars{\theta}
\,\right\vert_{\,\theta\ =\ 0}^{\,\theta\ \to\ \infty}
\\[3mm]&=\color{#66f}{\large{1 \over R}}
\end{align}
