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: = (2x3) / Sqrt(4xx^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 

Tags 
integral 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
a new discovery in integral calculus??for integral pros only  gen_shao  Calculus  2  July 31st, 2013 09:54 PM 
My found integral vs Wolfram Alpha's found integral  Obsessed_Math  Calculus  4  February 9th, 2012 04:18 PM 
integral of double integral in a region E  maximus101  Calculus  0  March 4th, 2011 02:31 AM 
Prove Lower Integral <= 0 <= Upper Integral  xsw001  Real Analysis  1  October 29th, 2010 07:27 PM 
integral of double integral in a region E  maximus101  Algebra  0  January 1st, 1970 12:00 AM 