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Vamos fazer uma substituição:

Usando a substituição na integral dada:

Vamos fatorar o polinômio (y⁴+1) usando o Lema de Gauss. Podemos ver que não ha raízes reais, logo vamos diretamente tentar fatorar em dois polinômios de grau 2:

Agora podemos usar o método de frações parciais. Olhando agora para a fração de dentro da integral:

Usando o que foi feito acima:

Resolveremos cada uma das integrais destacadas separadamente. Começando por I₁:

Agora, vamos calcular I₂:

Podemos voltar à expressão de I:

Agora, voltando à expressão em função de x, chegamos finalmente à resposta final:

Usando a substituição na integral dada:
Vamos fatorar o polinômio (y⁴+1) usando o Lema de Gauss. Podemos ver que não ha raízes reais, logo vamos diretamente tentar fatorar em dois polinômios de grau 2:
Agora podemos usar o método de frações parciais. Olhando agora para a fração de dentro da integral:
Usando o que foi feito acima:
Resolveremos cada uma das integrais destacadas separadamente. Começando por I₁:
Agora, vamos calcular I₂:
Podemos voltar à expressão de I:
Agora, voltando à expressão em função de x, chegamos finalmente à resposta final:
FelipeQueiroz:
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