# Thread: Integration with a square root

1. ## Integration with a square root

$\displaystyle \int_{1}^{4}\sqrt{t^2+4+t^{-2}}dt$

I've never seen an integral of this form, and I need help solving it. Am I expected to use tables?

The original problem asks for the length of the three dimensional curve given by r(t) = (2t, ln(t), t^2)

The answer in the back of the book is L = 14 + ln(4)

I set the arc length forumla up and arrived to this integral. Using the t89, I confirmed my setup results in the correct answer. However, I do not know how to arrive to that answer.

2. Originally Posted by 1005
$\displaystyle \int_{1}^{4}\sqrt{t^2+4+t^{-2}}dt$

I've never seen an integral of this form, and I need help solving it. Am I expected to use tables?

The original problem asks for the length of the three dimensional curve given by r(t) = (2t, ln(t), t^2)

The answer in the back of the book is L = 14 + ln(4)

I set the arc length forumla up and arrived to this integral. Using the t89, I confirmed my setup results in the correct answer. However, I do not know how to arrive to that answer.
is this an arc length problem? if so, what is the original function?

3. Originally Posted by 1005
$\displaystyle \int_{1}^{4}\sqrt{t^2+4+t^{-2}}dt$

I've never seen an integral of this form, and I need help solving it. Am I expected to use tables?

The original problem asks for the length of the three dimensional curve given by r(t) = (2t, ln(t), t^2)

The answer in the back of the book is L = 14 + ln(4)

I set the arc length forumla up and arrived to this integral. Using the t89, I confirmed my setup results in the correct answer. However, I do not know how to arrive to that answer.
Originally Posted by skeeter
is this an arc length problem? if so, what is the original function?

sdfsd

4. integral should be ...

$\displaystyle S = \int_1^4 \sqrt{4 + \frac{1}{t^2} + 4t^2} \, dt$

note that ...

$\displaystyle 4t^2 + 4 + \frac{1}{t^2} = \left(2t + \frac{1}{t}\right)^2$

can you finish?

5. Originally Posted by skeeter
integral should be ...

$\displaystyle S = \int_1^4 \sqrt{4 + \frac{1}{t^2} + 4t^2} \, dt$

note that ...

$\displaystyle 4t^2 + 4 + \frac{1}{t^2} = \left(2t + \frac{1}{t}\right)^2$

can you finish?
That's the same integral I wrote just with t^-2 instead of 1/t^2(the same thing). Nice factoring. Yes, I can solve that.

6. Originally Posted by 1005
That's the same integral I wrote just with t^-2 instead of 1/t^2(the same thing). Nice factoring. Yes, I can solve that.
No you left out the 4 in front of $\displaystyle t^2$ making the integral much more difficult