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?

Solution 1:

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

$$-\frac12 \cdot \frac23 (1+v)^{3/2} + C = -\frac13 \left (1+\frac1{x^2} \right )^{3/2}+C$$