<s>[INST] What is the fast way to evaluate the following integral: $\int{\frac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$? I am trying to evaluate the following integral:
$$\int{\dfrac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$$
I tried the trigonometric substitution: $u = \tan(x)$. Generally, The whole integral needs two substitutions: $u = \tan(x)$ then $v = \sin(u)$. In order to get rid of trigonometric functions, one needs to know that: $$\sin(\arctan(x))=\dfrac{x}{\sqrt{x^2+1}}$$
My question is: What is the fast substitution that leads to the answer without passing by the above steps?
 [/INST] For clarifying the sign of the antiderivative, I decide to integrate it by cases:
A. When $x>0,$
$$
\begin{aligned}
I &=\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} d x \\
&=-\frac{1}{2} \int \sqrt{1+\frac{1}{x^{2}}} d\left(1+\frac{1}{x^{2}}\right) \\
&=-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}+C\\&=-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C
\end{aligned}
$$
B. When $x<0,$ let $y=-x$, then $$
I=\int \frac{\sqrt{y^{2}+1}}{y^{4}}(-d y)=-\int \frac{\sqrt{y^{2}+1}}{y^{4}} d y=\frac{\left(y^{2}+1\right)^{2}}{3 y^{3}}+C =-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C
$$
Therefore we can conclude that
$$I=-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C.$$
</s>