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July 15th, 2007, 11:01 AM   #1
 
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Integral ... HELP PLEASE!!!

Can someone please help me with the below problem and show ALL steps to derive the answer ... if you can please include your substitution etc. ...

Problem:

Find the indefinite integral of:

= (2x-3) / Sqrt(4x-x^2) dx

THANKS A LOT FOR YOUR HELP!!!!

Miranda
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July 23rd, 2007, 12:11 PM   #2
 
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( 2x - 3 ) / sqrt ( 4x - x^2 ) = ( 2x - 4 + 1 ) / sqrt ( 4x - x^2 ) = [ ( 2x - 4 ) / sqrt ( 4x - x^2 ) ] + [ 1 / sqrt ( 4x - x^2 ) ]

I = integral [ ( 2x - 3 ) / sqrt ( 4x - x^2 ) ] dx = integral [ ( 2x - 4 ) / sqrt ( 4x - x^2 ) ] dx + integral [ 1 / sqrt ( 4x - x^2 ) ] dx = I' + I''

I' = integral [ ( 2x - 4 ) / sqrt ( 4x - x^2 ) ] dx
I'' = integral [ 1 / sqrt ( 4x - x^2 ) ] dx

I' = integral [ ( 2x - 4 ) / sqrt ( 4x - x^2 ) ] dx = - integral [ ( 4 - 2x ) / sqrt ( 4x - x^2 ) ] dx = - 2 . sqrt ( 4x - x^2 )

I'' = integral [ 1 / sqrt ( 4x - x^2 ) ] dx = integral [ 1 / sqrt ( 4 - 4 + 4x - x^2 ) ] dx = integral [ 1 / sqrt ( 4 - ( x - 2 ) ^ 2 ) ] dx
2u = x - 2
2du = dx
I'' = integral [ 1 / sqrt ( 4 - 4 u ^ 2 ) ] 2 . du = integral [ 1 / sqrt ( 1 - u ^ 2 ) ] du = arcsin ( u ) = arcsin [ ( x - 2 ) / 2 ]

I = I' + I'' = arcsin [ ( x - 2 ) / 2 ] - 2 . sqrt ( 4x - x^2 ) + K

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