integral from (sqrt[2]/3) to (2/3) of 1/(x*Sqrt[9x-1]) dx Thank you!
P pamw New member Joined Jan 10, 2007 Messages 8 Jan 10, 2007 #1 integral from (sqrt[2]/3) to (2/3) of 1/(x*Sqrt[9x-1]) dx Thank you!
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jan 11, 2007 #2 Try this: Let u=9x−1 and du=929x−1dx and 29du=19x−1dx and x=u2+19\displaystyle u=\sqrt{9x-1} \;\ and \;\ du=\frac{9}{2\sqrt{9x-1}}dx \;\ and \;\ \frac{2}{9}du=\frac{1}{\sqrt{9x-1}}dx \;\ and \;\ x=\frac{u^{2}+1}{9}u=9x−1 and du=29x−19dx and 92du=9x−11dx and x=9u2+1 There, now it'll fall together. Sorry 'bout that You should end up with: 2u2+1\displaystyle \frac{2}{u^{2}+1}u2+12 Don't forget to change your limits of integration accordingly.
Try this: Let u=9x−1 and du=929x−1dx and 29du=19x−1dx and x=u2+19\displaystyle u=\sqrt{9x-1} \;\ and \;\ du=\frac{9}{2\sqrt{9x-1}}dx \;\ and \;\ \frac{2}{9}du=\frac{1}{\sqrt{9x-1}}dx \;\ and \;\ x=\frac{u^{2}+1}{9}u=9x−1 and du=29x−19dx and 92du=9x−11dx and x=9u2+1 There, now it'll fall together. Sorry 'bout that You should end up with: 2u2+1\displaystyle \frac{2}{u^{2}+1}u2+12 Don't forget to change your limits of integration accordingly.
