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int dt/sqrt[3t-2t^2]...

`int dt/sqrt[3t-2t^2]`

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The correct Answer is:
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Let `I = int(dt)/(sqrt(3t-2t^(2)))=1/(sqrt(2))int(dt)/(sqrt(-(t^(2)-3/2t)))`
`=1/2int (dt)/(sqrt(-[(t^(2)-2.(1)/2.(3)/(2)t)+(3/4)^(2)-(3/4)^(2)]))`
`=1/(sqrt(2))int(dt)/(sqrt(-[(t-3/4)^(2)-(3/4)^(2)]))`
`= 1/(sqrt(2))int(dt)/(sqrt((3/4)^(2)-(t-(3)/(4))^(2)))`
`= 1/(sqrt(2))sin^(-1)((t-(3)/(4))/(3/4)) + C = 1/(sqrt(2))sin^(-1)((4t-3)/(3))+C`
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