Calcule as integrais usando funções Gamma e Beta
\int_{0}^{\infty}\left(1+x^{2}\right)^{2} \exp \left(-x^{2}\right) d x

Passo 1

O enunciado pede para calcular
\int_{0}^{\infty}\bigl(1+x^{2}\bigr)^{2}e^{-x^{2}}\,dx
usando propriedades das funções Gamma e Beta. Vamos abrir o quadrado, separar em integrais mais simples e aplicar a fórmula padrão
\int_{0}^{\infty}x^{n}e^{-x^{2}}\,dx = \frac{1}{2}\,\Gamma\!\bigl(\tfrac{n+1}{2}\bigr).

Passo 2

Expandimos o integrando:
\bigl(1+x^{2}\bigr)^{2} = 1 + 2x^{2} + x^{4}.
Logo
\int_{0}^{\infty}(1+x^{2})^{2}e^{-x^{2}}\,dx = \int_{0}^{\infty}e^{-x^{2}}\,dx + 2\int_{0}^{\infty}x^{2}e^{-x^{2}}\,dx + \int_{0}^{\infty}x^{4}e^{-x^{2}}\,dx.

Passo 3

Aplicamos a fórmula com n=0,2,4:
\int_{0}^{\infty}x^{n}e^{-x^{2}}\,dx = \frac{1}{2}\,\Gamma\!\bigl(\tfrac{n+1}{2}\bigr).

Passo 4

Calculando cada termo:
Para n=0:
\int_{0}^{\infty}e^{-x^{2}}\,dx = \tfrac{1}{2}\,\Gamma\!\bigl(\tfrac{1}{2}\bigr) = \tfrac{1}{2}\sqrt{\pi}.
Para n=2:
\int_{0}^{\infty}x^{2}e^{-x^{2}}\,dx = \tfrac{1}{2}\,\Gamma\!\bigl(\tfrac{3}{2}\bigr) = \tfrac{1}{2}\cdot\tfrac{1}{2}\sqrt{\pi} = \tfrac{\sqrt{\pi}}{4}.
Para n=4:
\int_{0}^{\infty}x^{4}e^{-x^{2}}\,dx = \tfrac{1}{2}\,\Gamma\!\bigl(\tfrac{5}{2}\bigr) = \tfrac{1}{2}\cdot\tfrac{3}{4}\sqrt{\pi} = \tfrac{3\sqrt{\pi}}{8}.

Passo 5

Somando as contribuições:
\tfrac{1}{2}\sqrt{\pi} +2\cdot\tfrac{\sqrt{\pi}}{4} +\tfrac{3\sqrt{\pi}}{8} = \frac{4\sqrt{\pi}}{8} + \frac{4\sqrt{\pi}}{8} + \frac{3\sqrt{\pi}}{8} = \frac{11\sqrt{\pi}}{8}.

Resposta

\int_{0}^{\infty}\bigl(1+x^{2}\bigr)^{2}e^{-x^{2}}\,dx = \frac{11\sqrt{\pi}}{8}.

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