How do I evaluate the integral $$\int \frac{\sqrt{9+x^{2}}}{x^{4}} dx ?$$
I found one method that is wonderful (see below). What other methods can be used?
Substitute $$z=\dfrac{1}{x^{2}}, \qquad d z=-\dfrac{2}{x^{3}} d x. $$ Then, \begin{aligned} I &=-\frac{1}{2} \int \frac{\sqrt{9+\frac{1}{z}}}{\frac{1}{\sqrt{z}}} d z \\ &=-\frac{1}{2} \int \sqrt{9 z+1} d z \\ &=-\frac{(9 z+1)^{\frac{3}{2}}}{27}+C\\ &=-\frac{\left(9+x^{2}\right)^{\frac{3}{2}}}{27 x^{3}}+C \end{aligned}
Probably this question is essentially a duplicate, but I cannot find another instance with a cursory search.
We may as well produce a more general result by replacing $9$ with $a^2$ for some $a > 0$.
Method 1 Because $x$ appears in a radical only in $\sqrt{a^2 + x^2}$, the standard method is to apply the trigonometric substitution $$x = a \tan \theta, \qquad dx = a \sec^2 \theta \,d\theta ,$$ which transforms the integral into $$\frac{1}{a^2} \int \frac{\cos \theta \,d\theta}{\sin^4 \theta},$$ which can readily be evaluated with a single $u$-substitution.
Method 2 Yet another option is the hyperbolic trigonometric substitution $$x = a \sinh t, \qquad dx = a\cosh t\,dt ,$$ which transforms the integral to $$\frac{1}{a^2} \int \coth^2 t \operatorname{csch}^2 t\,dt,$$ which can again be evaluated using a single $u$-substitution.
Method 3 The single substitution $$v = \frac{x}{\sqrt{a^2 + x^2}}, \qquad dv = \frac{a^2 \,dx}{(a^2 + x^2)^{3 / 2}},$$ transforms the integral to $$\frac{1}{a^2} \int \frac{dv}{v^4} ,$$ but this substitution is harder to see than the previous ones and arguably no faster.
Method 4 Another single substitution, $$x = \frac{2 a w}{1 - w^2} , \qquad dx = \frac{2 a (w^2 + 1)}{(w^2 - 1)^2} dw ,$$ transforms the integral to $$\frac{1}{8 a^2} \int \left(w^2 + 1 - w^{-2} - w^{-4}\right) dt ,$$ but that choice makes back-substitution a little unpleasant.