Integral (lower limit is 1, upper limit 9): (x-1)/(x)^1/2 how do you do it ?
H HardikT New member Joined Nov 10, 2009 Messages 2 Nov 10, 2009 #1 Integral (lower limit is 1, upper limit 9): (x-1)/(x)^1/2 how do you do it ?
pka Elite Member Joined Jan 29, 2005 Messages 11,976 Nov 10, 2009 #2 \(\displaystyle \frac{{x - 1}}{{\sqrt x }} = x^{\frac{1}{2}} - x^{ - \frac{1}{2}}\)
