<s>[INST] 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
 [/INST] 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
$$
\begin{align}
\int\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y
&=\frac1R\int\cos(\theta)\,\mathrm{d}\theta\\
&=\frac1R\sin(\theta)+C\\
&=\frac1R\frac{y}{\sqrt{R^2+y^2}}+C
\end{align}
$$
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