How can $\int \frac{dx}{(x+a)^2(x+b)^2}$ be found? Could you please suggest any hints or methods for solving $\int \frac{dx}{(x+a)^2(x+b)^2}$. I have used partial fractions to solve this integral but it is too long and complex solution. I'd like to know a simpler solution.
EDIT: $a\not= b$
 A: Hint:
$$\frac{1}{(x+a)^2 (x+b)^2 } = \frac{-2}{(a - b)^3 (b + x) } + \frac{1}{(a - b)^2 (b + x)^2} + \frac{2}{(a - b)^3 (a + x)} + \frac{1}{(a - b)^2 (a + x)^2}$$
by partial fraction decomposition.
A: HINT:
Use  $a-b=x+a-(x+b)$ and $$(a-b)^2=\{(x+a)-(x+b)\}^2=\cdots$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \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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\begin{align}
&\int{\dd x \over \pars{x + a}^{2}\pars{x + b}^{2}}  =
{\partial^{2} \over \partial a\,\partial b}
\int{\dd x \over \pars{x + a}\pars{x + b}} =
{\partial^{2} \over \partial a\,\partial b}
\int\pars{{1 \over x + a} - {1 \over x + b}}{\dd x \over b - a}
\\[5mm] = &\ 
{\partial^{2} \over \partial a\,\partial b}\bracks{%
\ln\pars{x + a \over x + b}\,{1 \over b - a}}
\\[5mm] = &\
\bbx{\ds{\bracks{2\ln\pars{x + a \over x + b} -
\pars{a - b}\,{2x + a + b \over \pars{x + a}\pars{x + b}}}
{1 \over \pars{a - b}^{3}} + \pars{~\mbox{a constant}~}}}
\end{align}
