integral, from sqrt[2]/3 to 2/3, of 1/(x sqrt[9x-1])dx
- Thread starter pamw
- Start date
Try this:
Let \(\displaystyle u=\sqrt{9x-1} \;\ and \;\ du=\frac{9}{2\sqrt{9x-1}}dx \;\ and \;\ \frac{2}{9}du=\frac{1}{\sqrt{9x-1}}dx \;\ and \;\ x=\frac{u^{2}+1}{9}\)
There, now it'll fall together. Sorry 'bout that
You should end up with:
\(\displaystyle \frac{2}{u^{2}+1}\)
Don't forget to change your limits of integration accordingly.
Let \(\displaystyle u=\sqrt{9x-1} \;\ and \;\ du=\frac{9}{2\sqrt{9x-1}}dx \;\ and \;\ \frac{2}{9}du=\frac{1}{\sqrt{9x-1}}dx \;\ and \;\ x=\frac{u^{2}+1}{9}\)
There, now it'll fall together. Sorry 'bout that
You should end up with:
\(\displaystyle \frac{2}{u^{2}+1}\)
Don't forget to change your limits of integration accordingly.