Melody

That link does not work.

x^11*x^7 = x^18

the answer is 18

It would be nice to get feedback though 

I answered on the other post.

Sorry, I know i asked you to put it on a new post but I then got interested and answered it on the original  :)

If anyone wants to show guest how to do it in a more traditional way.

That is fine :)

 ∫5 3 (1/(x^2-6x+13))dx

\(\displaystyle\int_3^{5}\frac{1}{\sqrt{x^2-6x+13}}dx\\ =\displaystyle\int_3^{5}\frac{1}{\sqrt{x^2-6x+9+4}}dx\\ =\displaystyle\int_3^{5}\frac{1}{\sqrt{2^2+(x-3)^2}}dx\\\)

I did this one in a very similar fashion.

I used this triangle

See if you can take it from there   

deleted

I probably do these a little differently from most people.

I cannot remember the integrals so I so them more from scratch.

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\)

I am going to use this triangle.

\(cos\theta=\sqrt{1-9x^2}\\ sin\theta=3x\\ x=\frac{sin\theta}{3}\\ \frac{dx}{d\theta}=\frac{cos\theta}{3}\\ dx=\frac{cos\theta}{3}d\theta\)

also

\(when \;x=0  \\ sin\theta=0\\ \theta=0\\~\\ when\;x=1/6\\ sin\theta=3*(1/6)=1/2\\ \theta = \frac{\pi}{6}\)

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\\ =\displaystyle\int_0^{\frac{1}{6}}\frac{1}{cos\theta}dx\\ =\displaystyle\int_0^{\frac{\pi}{6}}\frac{1}{cos\theta}\frac{cos\theta}{3}d\theta\\ =\displaystyle\int_0^{\frac{\pi}{6}}\frac{1}{3}d\theta\\ =\left[\frac{\theta}{3}\right]_0^{\frac{\pi}{6}}\\ =\frac{\pi}{18} \)

∫1/6 0 (1/(sqrt(1-9x^2))dx

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\)

Here is a site that should help

Well you'd add 8 for part 1

But the final answer would be bigger by 8*24

Note: I have not checked all you working but your logic seems good.