Is there an identity to solve for: $ \int \frac{dx}{(x^2+ \alpha^2)^2} $ Is there an identity to solve for: $ \int \frac{dx}{(x^2+ \alpha^2)^2} $
Thanks in advance and more power.
 A: We wish to solve
$$
I=\int \frac{dx}{(x^2+ \alpha^2)^2}
$$
We can use a hyperbolic substitution of the form $x=\alpha\sinh \phi, dx=\alpha\cosh \phi d\phi$
$$
I=\int \frac{\alpha\cosh \phi d\phi}{(\alpha^2(\sinh^2 \phi+1))^{2}}=\frac{1}{\alpha^3}\int \frac{\cosh \phi d\phi}{\cosh^4 \phi}=\frac{1}{\alpha^3}\int \frac{d\phi}{\cosh^3 \phi}=\frac{1}{\alpha^3}\left( \arctan\big(\tanh \frac{\phi}{2}\big)+\frac{1}{2}\text{sech} \phi \tanh \phi + \mathcal{C}\right).
$$
Changing back to our original variable x using $\phi=\sinh^{-1}(x/\alpha)$ we obtain 
$$
I=\frac{ \frac{\alpha x}{\alpha^2+x^2}+\arctan(\frac{x}{\alpha})}{2\alpha^3}+\mathcal{C}.
$$
Note, you may find these useful for the future
$$
\tanh \big(\sinh^{-1}\frac{x}{\alpha}\big)=\frac{x}{\sqrt{\alpha^2+x^2}}, \ \ \text{sech}\big({\sinh^{-1} \frac{x}{\alpha}\big)= \frac{\alpha}{\sqrt{\alpha^2+x^2}}}, \ \alpha >0.
$$
A: Hint
Integrate by parts
$$\int\frac{dx}{x^2+\alpha^2}=\frac{x}{x^2+\alpha^2}+2\int\frac{x^2\color{red}{+\alpha^2-\alpha^2}}{(x^2+\alpha^2)^2}dx$$
Can you take it from here?
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#00f}{\large\int{\dd x  \over \pars{x^{2} + a^{2}}^{2}}}=
-\,{1 \over 2a}\,\partiald{}{a}\bracks{\int{\dd x  \over x^{2} + a^{2}}}
=-\,{1 \over 2a}\,\partiald{}{a}\bracks{{1 \over a}
\int{\dd\pars{x/a}  \over \pars{x/a}^{2} + 1}}
\\[3mm]&=-\,{1 \over 2a}\,\partiald{}{a}\bracks{{1 \over a}\,\arctan\pars{x \over a}}
=-\,{1 \over 2a}\bracks{-\,{1 \over a^{2}}\,\arctan\pars{x \over a}
+ {1 \over a}\,{-\pars{x/a^{2}} \over \pars{x/a}^{2} + 1}}
\\[3mm]&=\color{#00f}{\large%
{1 \over 2a^{3}}\bracks{\arctan\pars{x \over a}
+ {a\,x \over x^{2} + a^{2}}}} + \mbox{a constant}
\end{align}
A: Using Trigonometric substitution, set $$x=\alpha\tan\theta$$
Then  use Double Angle formula $\displaystyle\displaystyle\cos2\theta=2\cos^2\theta-1\iff\cos^2\theta=\frac{1+\cos2\theta}2$
Now, $\displaystyle\int2\cos2\theta\ d\theta=\sin2\theta+K=\frac{2\tan\theta}{1+\tan^2\theta}+K$ where $K$ is an arbitrary constant 
