My Math Forum Integral ... HELP PLEASE!!!
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 July 15th, 2007, 11:01 AM #1 Newbie   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 Newbie   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

 Tags integral

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# integral of (4x^2-x-2)/(x^3 x^2)

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post gen_shao Calculus 2 July 31st, 2013 09:54 PM Obsessed_Math Calculus 4 February 9th, 2012 04:18 PM maximus101 Calculus 0 March 4th, 2011 02:31 AM xsw001 Real Analysis 1 October 29th, 2010 07:27 PM maximus101 Algebra 0 January 1st, 1970 12:00 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top