[Cambridge/2003 modified] Use the substitution $x = \tan \theta$ to find
$\int \frac{1 - x^2}{(1 + x^2)^2} \, dx.$

1. Find the differential $dx$ in terms of $d\theta$.

$x = \tan \theta \implies \frac{dx}{d\theta} = \sec^2 \theta$
$dx = \sec^2 \theta \, d\theta$

2. Substitute $x$ and $dx$ into the integral and simplify the denominator.

Substitute into the integral:
$\int \frac{1 - \tan^2 \theta}{(1 + \tan^2 \theta)^2} \sec^2 \theta \, d\theta$
Using the identity $1 + \tan^2 \theta = \sec^2 \theta$:
$= \int \frac{1 - \tan^2 \theta}{(\sec^2 \theta)^2} \sec^2 \theta \, d\theta$
$= \int \frac{1 - \tan^2 \theta}{\sec^2 \theta} \, d\theta$

3. Simplify the integrand using trigonometric identities.

$= \int \left( \frac{1}{\sec^2 \theta} - \frac{\tan^2 \theta}{\sec^2 \theta} \right) d\theta$
$= \int \left( \cos^2 \theta - \frac{\sin^2 \theta / \cos^2 \theta}{1 / \cos^2 \theta} \right) d\theta$
$= \int (\cos^2 \theta - \sin^2 \theta) \, d\theta$
Using the double angle identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$:
$= \int \cos 2\theta \, d\theta$

4. Evaluate the integral with respect to $\theta$.

$= \frac{1}{2} \sin 2\theta + C$

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

Using the double angle identity $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \sin \theta \cos \theta + C$
From $x = \tan \theta$, we can construct a right-angled triangle with opposite $= x$ and adjacent $= 1$. The hypotenuse is $\sqrt{1+x^2}$.
$\sin \theta = \frac{x}{\sqrt{1+x^2}} \quad \text{and} \quad \cos \theta = \frac{1}{\sqrt{1+x^2}}$
Substituting these back:
$= \left(\frac{x}{\sqrt{1+x^2}}\right)\left(\frac{1}{\sqrt{1+x^2}}\right) + C$
$= \frac{x}{1+x^2} + C$