My Math Forum Integral ... HELP PLEASE!!!

 Calculus Calculus Math Forum

 July 15th, 2007, 11:01 AM #1 Joined: Jul 2007 Posts: 1 Thanks: 0 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
 July 23rd, 2007, 12:11 PM #2 Joined: Jul 2007 Posts: 2 Thanks: 0 ( 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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# X-1/square root of x^2-4x 3

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