Kindly provide the stepwise solution for this integral. $$\int_{-b}^b\int_{-a}^a\frac{1}{(x^2+y^2+h^2)^{3/2}}dxdy$$
First I put x=$\sqrt{y^2+h^2}tan\theta$ and arrive at: 
$$\int_{-b}^b\frac{2a}{(y^2+h^2)(y^2+h^2+a^2)^{1/2}}dy$$
Can you please tell me what to do after this?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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Note that
  $\ds{\pars{~\mbox{with}\ a\,,\ b > 0~}}$ and
  $\ds{\pars{p \equiv {a \over \verts{h}} > 0\,,\
q \equiv {b \over \verts{h}} > 0}}$:

$$
\int_{-a}^{a}\int_{-b}^{b}{\dd x\,\dd y \over
\pars{x^{2} + y^{2} + h^{2}}^{3/2}} =
{4 \over \verts{h}}\bbox[5px,#ffd]{\ds{\int_{0}^{q}\int_{0}^{p}{\dd x\,\dd y \over
\pars{x^{2} + y^{2} + 1}^{3/2}}}}
$$

\begin{align}
&\bbox[5px,#ffd]{\ds{\int_{0}^{q}\int_{0}^{p}{\dd x\,\dd y \over
\pars{x^{2} + y^{2} + 1}^{3/2}}}} \,\,\,\stackrel{y\ \mapsto\ {1/y}}{=}\,\,\,
\int_{0}^{p}\int_{1/q}^{\infty}{y\,\dd y \over
\bracks{\pars{x^{2} + 1}y^{2} + 1}^{3/2}}\,\dd x
\\[5mm] \stackrel{y^{2}\ \mapsto\ y}{=} &\
{1 \over 2}\int_{0}^{p}\int_{1/q^{2}}^{\infty}{\dd y \over
\bracks{\pars{x^{2} + 1}y + 1}^{3/2}}\,\dd x =
-\int_{0}^{p}{1 \over x^{2} + 1}\left.%
{1 \over \bracks{\pars{x^{2} + 1}y + 1}^{1/2}}
\right\vert_{\ y\ =\ 1/q^{2}}^{\ y\ \to\ \infty}\,\dd x
\\[5mm] = &\
q\int_{0}^{p}{1 \over x^{2} + 1}
{1 \over \bracks{\pars{x^{2} + 1} + q^{2}}^{1/2}}\,\dd x
\,\,\,\stackrel{x\ \mapsto\ {1/x}}{=}\,\,\,
q\int_{1/p}^{\infty}
{x\,\dd x \over \pars{x^{2} + 1}\bracks{\pars{q^{2} + 1}x^{2} + 1}^{1/2}}
\\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,&
{1 \over 2}\,q\int_{1/p^{2}}^{\infty}
{\dd x \over \pars{x + 1}\bracks{\pars{q^{2} + 1}x + 1}^{1/2}}
\end{align}


With the change of variable
  $\ds{x \equiv {t^{2} - 1 \over q^{2} + 1}}$:

\begin{align}
&\bbox[5px,#ffd]{\ds{\int_{0}^{q}\int_{0}^{p}{\dd x\,\dd y \over
\pars{x^{2} + y^{2} + 1}^{3/2}}}} =
q\int_{\root{p^{2} + q^{2} + 1}/p}^{\infty}{\dd t \over t^{2} + q^{2}} =
{\pi \over 2} - \arctan\pars{\root{p^{2} + q^{2} + 1} \over pq}
\\[5mm] = &\
\arctan\pars{pq \over \root{p^{2} + q^{2} + 1}}
\end{align}

$$
\bbx{\ds{\int_{-a}^{a}\int_{-b}^{b}{\dd x\,\dd y \over
\pars{x^{2} + y^{2} + h^{2}}^{3/2}} =
{4 \over h}\,\arctan\pars{ab/h \over \root{a^{2} + b^{2} + h^{2}}}}}
$$
A: If we substitute
$$y=\sqrt{a^2+h^2}\tan u$$
$$dy=\sqrt{a^2+h^2}\sec^2u \text{ }du$$
We get the integral
$$\int \frac{2a(\sqrt{a^2+h^2}\sec^2u)}{((a^2+h^2)\tan^2u+h^2)((a^2+h^2)\tan^2u+a^2+h^2)^{1/2}}du$$
Remembering that $\tan^2x+1=\sec^2x$, we get
$$\int \frac{2a\sqrt{a^2+h^2}\sec^2u}{((a^2+h^2)\sec^2u -h^2)((a^2+h^2)\sec^2u)^{1/2}}du$$ 
Now you can simplify quite a bit.
Hope this helps!
