<s>[INST] How to simplify the integration? If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows the $\tan^{-1}x$ but How to simplify if we have
$$\int \frac{1}{(1+x^2)^2}dx$$
 [/INST] Define $f(\alpha)=\int\dfrac{dx}{\alpha+x^2}=\frac{1}{\sqrt\alpha}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)$ 
You can recover the original integral by putting $\alpha=1$ later, 
Now, differentiating both sides.
$f'(\alpha)=\int\dfrac{-dx}{(\alpha+x^2)^2}=\frac{-1}{2(\alpha)^{3/2}}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)+\frac{1}{(\alpha)^{1/2}}\frac{1}{1+\frac{x^2}{\alpha}}\frac{-x}{2(\alpha)^{3/2}}$ 
$\int\dfrac{dx}{(\alpha+x^2)^2}=\frac{1}{2(\alpha)^{3/2}}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)+\frac{1}{(\alpha)^{1/2}}\frac{1}{1+\frac{x^2}{\alpha}}\frac{x}{2(\alpha)^{3/2}}$
Put $\alpha=1$
$$\int\dfrac{dx}{(1+x^2)^2}=\frac{1}{2}\tan^{-1}x+\frac{1}{1+x^2}.\frac{x}{2}$$ and of course the remaining $C$
And there you have your integral found by differentiating only.^^
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