By using the substitution $x = \cot \theta$, for $0 < \theta < \frac{\pi}{4}$, find $\displaystyle \int \frac{1}{x^2\sqrt{1+x^2}}\,\mathrm{d}x$.

1. Find the derivative of the substitution to replace $\mathrm{d}x$.

Given the substitution $x = \cot \theta$, differentiating with respect to $\theta$ gives:
$\frac{\mathrm{d}x}{\mathrm{d}\theta} = -\csc^2 \theta$
$\mathrm{d}x = -\csc^2 \theta \,\mathrm{d}\theta$

2. Substitute $x$ and $\mathrm{d}x$ into the integral and simplify using trigonometric identities.

Substitute into the original integral:
$\int \frac{1}{x^2\sqrt{1+x^2}} \,\mathrm{d}x = \int \frac{1}{\cot^2 \theta \sqrt{1+\cot^2 \theta}} (-\csc^2 \theta) \,\mathrm{d}\theta$
Using the identity $1+\cot^2 \theta = \csc^2 \theta$ and knowing $\csc \theta > 0$ for $0 < \theta < \frac{\pi}{4}$:
$= \int \frac{-\csc^2 \theta}{\cot^2 \theta \csc \theta} \,\mathrm{d}\theta$
$= \int -\frac{\csc \theta}{\cot^2 \theta} \,\mathrm{d}\theta$
Expressing the integrand in terms of sines and cosines:
$= \int -\frac{1/\sin \theta}{\cos^2 \theta / \sin^2 \theta} \,\mathrm{d}\theta$
$= \int -\frac{\sin \theta}{\cos^2 \theta} \,\mathrm{d}\theta$
$= \int -\sec \theta \tan \theta \,\mathrm{d}\theta$

3. Evaluate the simplified integral with respect to $\theta$.

Evaluating the standard trigonometric integral:
$= -\sec \theta + C$

4. Express the final answer in terms of the original variable $x$.

To express the answer in terms of $x$, use the given substitution $x = \cot \theta = \frac{x}{1}$.
Consider a right-angled triangle where the adjacent side to $\theta$ is $x$ and the opposite side is $1$.
By Pythagoras' theorem:
$\text{Hypotenuse} = \sqrt{x^2 + 1^2} = \sqrt{1+x^2}$
Therefore, $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{\sqrt{1+x^2}}{x}$.
Substituting back:
$-\sec \theta + C = -\frac{\sqrt{1+x^2}}{x} + C$