A akistar New member Joined May 15, 2014 Messages 1 May 15, 2014 #1 Need help on this one..really stuck this time:-( Evaluate Definite Integral
stapel Elite Member Joined Feb 4, 2004 Messages 16,543 May 15, 2014 #2 akistar said: Need help on this one..really stuck this time:-( Evaluate Definite Integral \(\displaystyle \displaystyle{\int_{-2}^{-1}}\, \frac{-3x^2\, +\, 6x\, -\, 2}{3x^2\, -\, 2x}\, dx\) Click to expand... One possible first step might be to do the long polynomial division, to convert this "fraction" into "mixed-number" form. Then integrate the terms....
akistar said: Need help on this one..really stuck this time:-( Evaluate Definite Integral \(\displaystyle \displaystyle{\int_{-2}^{-1}}\, \frac{-3x^2\, +\, 6x\, -\, 2}{3x^2\, -\, 2x}\, dx\) Click to expand... One possible first step might be to do the long polynomial division, to convert this "fraction" into "mixed-number" form. Then integrate the terms....
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,760 May 15, 2014 #3 akistar said: Need help on this one..really stuck this time:-( Evaluate Definite Integral View attachment 4114 Click to expand... \(\displaystyle \frac{-3x^2+ 6x- 2}{3x^2- 2x}= -1+ \frac{4x+ 2}{x(3x- 2)}\) Use "partial fractions" to integrate the remaining fraction.
akistar said: Need help on this one..really stuck this time:-( Evaluate Definite Integral View attachment 4114 Click to expand... \(\displaystyle \frac{-3x^2+ 6x- 2}{3x^2- 2x}= -1+ \frac{4x+ 2}{x(3x- 2)}\) Use "partial fractions" to integrate the remaining fraction.