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?

We can simplify the expression $\frac{\sqrt{x^2+1}}{x^4}$, assuming that $x>0$, as follows:
$$\frac{\sqrt{x^2+1}}{x^4} = \frac{\sqrt{1+\frac1{x^2}}}{x^3} = \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}}.$$
(We have multiplied both numerator and denominator by $\frac1x$.)
After this simplification $t=\frac1x$ seems as a reasonable substitution. It leads to
$$\newcommand{\dd}{\; \mathrm{d}}\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x = 
\begin{vmatrix} x=\frac1t & \dd x = -\frac{\dd t}{t^2} \\ t=\frac1x & \dd t = -\frac{\dd x}{x^2}\end{vmatrix} = 
-\int t\sqrt{1+t^2} \dd t.$$
The last integral is rather simple.

 Hint: Try the substitution $u=1+t^2$.

Another reasonable substitution seems to be $t=\frac1{x^2}$, which leads to
$$\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x = 
\begin{vmatrix} t=\frac1{x^2} \\ \dd t=-\frac{\dd x}{x^3}\end{vmatrix} = -\int \sqrt{1+t} \dd t.$$
