Para resolver a integral \( \int_{0}^{1} \frac{1}{(x^2 + 1)^2} \, dx \), podemos usar a fórmula: \[ \int \frac{1}{(x^2 + a^2)^2} \, dx = \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a} \arctan\left(\frac{x}{a}\right) + C \] Neste caso, \( a = 1 \). Assim, a integral se torna: \[ \int \frac{1}{(x^2 + 1)^2} \, dx = \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan(x) + C \] Agora, avaliamos a integral definida de \( 0 \) a \( 1 \): \[ \left[ \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan(x) \right]_{0}^{1} \] Calculando em \( x = 1 \): \[ \frac{1}{2(1^2 + 1)} + \frac{1}{2} \arctan(1) = \frac{1}{4} + \frac{\pi}{8} \] Calculando em \( x = 0 \): \[ \frac{0}{2(0^2 + 1)} + \frac{1}{2} \arctan(0) = 0 \] Portanto, a integral é: \[ \frac{1}{4} + \frac{\pi}{8} - 0 = \frac{1}{4} + \frac{\pi}{8} \] Assim, a resposta correta é: \[ \int_{0}^{1} \frac{1}{(x^2 + 1)^2} \, dx = \frac{1}{4} + \frac{\pi}{8} \]