To integrate:

$I = \int \frac{1 - x^2}{(1+x^2)^2} \, dx$

Step 1: Use Substitution

Let:
$u = 1 + x^2 \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$

Step 2: Express the Integrand

Rewrite $1 - x^2$:
$1 - x^2 = 1 - (u - 1) = 2 - u$

The integral becomes:
$I = \int \frac{2 - u}{u^2} \cdot \frac{du}{2} = \int \frac{2 - u}{2u^2} \, du$

Step 3: Split the Terms

$I = \int \left(\frac{2}{2u^2} - \frac{u}{2u^2}\right) du$

Simplify:
$I = \int \left(\frac{1}{u^2} - \frac{1}{2u}\right) du$

Step 4: Integrate Each Term

$I = \int u^{-2} \, du - \frac{1}{2} \int u^{-1} \, du$

$I = \frac{-1}{u} - \frac{1}{2} \ln|u|$

Step 5: Substitute Back $u = 1 + x^2$

$I = \frac{-1}{1+x^2} - \frac{1}{2} \ln(1+x^2) + C$

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Final Answer:
$I = \frac{-1}{1+x^2} - \frac{1}{2} \ln(1+x^2) + C$

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Would you like further explanation on any of the steps?

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Related Problems:

1. Integrate $\frac{x^2}{(1+x^2)^2}$
2. Evaluate $\int \frac{1}{1+x^2} dx$
3. Integrate $\int \frac{x}{(1+x^2)^2} dx$
4. Find the derivative of $f(x) = \frac{-1}{1+x^2} - \frac{1}{2} \ln(1+x^2)$
5. Evaluate $\int \frac{1 - x^4}{(1+x^2)^3} dx$

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Tip:

When integrating rational functions, consider substitution with terms from the denominator or factor the numerator to simplify the expression.