I need help with an integral. It's part of an arc-length problem I'm working with but am stuck. I have the answer if that helps but am not sure how to get there.

Integral from 0 to 2 of the square root of (t^2 + 16t +16) dt

The answer in the back is (10root13 -16) - 24 ln((5+root13)/6).

would really appreciate any advice. thanks.

Question

I need help with an integral.  It's part of an arc-length problem I'm working with but am stuck.  I have the answer if that helps but am not sure how to get there.

Integral from 0 to 2 of the square root of (t^2 + 16t +16) dt

The answer in the back is (10root13 -16) - 24 ln((5+root13)/6).

would really appreciate any advice.  thanks.

anonymous · Answer

$$\int\limits_{0}^{2}\sqrt{t ^{2}+16t+16}dt$$

anonymous · Answer

Answer:$$(10\sqrt{13}-16)-24\ln ((5+\sqrt{13})/6)$$

anonymous · Answer

Have you tried completing the square under the radical?

anonymous · Answer

so that would give me the square root of (t+8)^2 -48, right?  then a trig substitution or something?

anonymous · Answer

Yes, just make sure you make the right sub.

anonymous · Answer

ok, cool.  will give it a shot.  thanks!