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∫√x2x+3dx
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Solution
√x−√32arctan(√23√x)+C
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Solve by:
One step at a time
∫√x2x+3dx
Aplicar integração por substituição:∫2u22u2+3du
=∫2u22u2+3du
Remover a constante: ∫a·f(x)dx=a·∫f(x)dx
=2·∫u22u2+3du
Divisão longa u22u2+3:12−32(2u2+3)
=2·∫12−32(2u2+3)du
Aplicar a regra da soma: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx