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integral of 1/square root of 16+4x-2x^2 times dx This is a calc 2 question think its trig substitution helppppp
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\[\int\limits_{}^{}1/\sqrt{16+4x-2x^2}\]
i made the bottom as \[\sqrt{-2(x-1)^2+1}\] then u=x-1 so gotta be \[\int\limits_{}^{}1/\sqrt{18-2u^2}\]
oh god thanx prony
*rony
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np =)
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