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?
 A: Rewrite as
$$\int dx \, x \frac1{x^4} \sqrt{1+\frac1{x^2}} = \frac12 \int du \frac1{u^2} \sqrt{1+\frac1{u}} = -\frac12 \int dv \sqrt{1+v}$$
In the above, $u=x^2$ and $v=1/u$.  Thus, the antiderivative is
$$-\frac12 \cdot \frac23 (1+v)^{3/2} + C = -\frac13 \left (1+\frac1{x^2} \right )^{3/2}+C$$
A: With $K=\int\frac{\sqrt{x^2+1}}{x^2}dx $, integrate by parts as follows
\begin{align}
\int\frac{\sqrt{x^2+1}}{x^4}dx
=& \int \frac{(x^2+1)^{3/2}}{x^4}-\frac{\sqrt{x^2+1}}{x^2}\ dx\\
=&\int (x^2+1)^{3/2}\ d\left(-\frac1{3x^3}\right)-K\\
\overset{ibp}=&\ -\frac{(x^2+1)^{3/2}}{3x^3}+K-K
\end{align}
A: 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
$$\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x = 
\begin{vmatrix} t=\frac1{x^2} \\ \dd t=-\frac{\dd x}{x^3}\end{vmatrix} = -\int \sqrt{1+t} \dd t.$$
A: 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
$$I=-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C.$$
