Evaluate the integral: I = integral (x²)/((1+x²)²) dx

Evaluate the integral: $$ I = \int \frac{x^2}{(1+x^2)^2}\,dx. $$

Step 1 · Rewrite the numerator

Since $x^2=(1+x^2)-1$:

$$ \frac{x^2}{(1+x^2)^2} = \frac{1+x^2}{(1+x^2)^2} - \frac{1}{(1+x^2)^2} = \frac{1}{1+x^2} - \frac{1}{(1+x^2)^2}. $$

Step 2 · Integrate each piece

The first piece is the standard arctangent integral:

$$ \int \frac{dx}{1+x^2} = \arctan x. $$

The second piece uses the standard reduction result:

$$ \int \frac{dx}{(1+x^2)^2} = \frac{x}{2(1+x^2)} + \frac12\arctan x. $$

Step 3 · Combine

$$ I = \arctan x - \left(\frac{x}{2(1+x^2)} + \frac12\arctan x\right) + C = \frac12\arctan x - \frac{x}{2(1+x^2)} + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$.
• The identity $x^2=(1+x^2)-1$ is the standard trick for reducing $x^2/(1+x^2)^n$ to a difference of two simpler $1/(1+x^2)^k$ integrals, avoiding a trigonometric substitution entirely.
• The result can be checked by differentiating $\tfrac12\arctan x-\tfrac{x}{2(1+x^2)}$: the quotient-rule derivative of the second term is $-\tfrac{(1+x^2)-2x^2}{2(1+x^2)^2}=\tfrac{x^2-1}{2(1+x^2)^2}$, and adding $\tfrac{1}{2(1+x^2)}=\tfrac{1+x^2}{2(1+x^2)^2}$ gives $\tfrac{2x^2}{2(1+x^2)^2}=\tfrac{x^2}{(1+x^2)^2}$, matching the integrand.