In an MIT OOCW video the prof has the integral of $\frac{1}{(x^2+4)^3}$ where $x=2tan\,u$

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In this video, the prof has used $x=a\tan u$ so $x=2\tan u$ for the integral of $\cfrac{dx}{(x^2+4)^3}$.

But I thought that could only be used with square roots of this case as in $\displaystyle \int\cfrac{dx}{\sqrt{x^2+4}}$

And without the square root then aren't you dealing with arctan?

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You can apply any substitution to any integral; you may have seen this particular one recommended for integrals of the form you mention, but that doesn't mean you can't use it elsewhere as well.

Of course, to use a substitution, you need to satisfy the hypotheses of whatever theorem you have that allows substitutions, but in practice, we often proceed by ignoring those, getting a result, and then checking that the derivative works out as we hoped (rather than losing a sign, etc.)