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Re: [obm-l] Integral
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] Integral
- From: "saulo nilson" <saulo.nilson@xxxxxxxxx>
- Date: Sat, 12 May 2007 13:25:50 -0300
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vc feza substituiçao errada
e^3x=u
du=3e^x^2*dx
e a integral se resume a
integral1/3*1/raiz(u^2+1) du
essa integral e facil acho que da
coshv=u
senhvdv=du
inte1/3 *senhvdv/senhv=1/3*intdv=v/3
voltando em x
arccoshe^3x/3 (1,00)
On 5/5/07, Marcus Aurélio <marcusaurelio80@globo.com> wrote:
Alguem sabe como resolver essa integral?
integral de 1 a mais infinito de e^2x sobre raiz quadrada de e^6x +1
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