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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

We got the formula \(\displaystyle \int_a^b\sqrt{1+[f'(x)]^2}\)

and \(\displaystyle f'(x)=\frac{x}{36}-\frac{9}{x} <=> \frac{x^2-324}{36x}\)

so now we got \(\displaystyle \int_9^{9e}\sqrt{1+(\frac{x^2-324}{36x})^2}\)

we can rewrite that as \(\displaystyle \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}\)

then do integration by part on it well I simply rewrite the function more too \(\displaystyle \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}\) and set \(\displaystyle u=1298x^2+(x^2-324)^2\) so we got \(\displaystyle dv=1296x^{-2}\) so our \(\displaystyle du=2592x+2x^2-648+x\) and \(\displaystyle v=\frac{1296x^{-3}}{-3}\)Is this correct?