Question

How to find the antiderivative of:

Answer 1

\[\int\limits_{1}^{x}\sqrt{1+(t^2-(1/(4t^2))^2} dt\]

Answer 2

I'm sorry it's (t^2 -(1/(4t^2)))^2

Answer 3

\[\large \int\limits_1^x \sqrt{1+\left(t^2-\frac{1}{4t^2}\right)^2}\]

It's a little hard to read the way you've written it out.
Is this correct?

Answer 4

Woops I forgot my dt :D whatev.

Answer 5

lol yes! Sorry for the confusion :)

Answer 6

Mmk so let's work from the inside out.\[\large t^2-\frac{1}{4t^2}\qquad = \qquad \frac{4t^4-1}{4t^2}\]I got a common denominator and combined them.

Answer 7

Now we have to square this..? Hmm

Answer 8

Squaring that term gives us this,
\[\large \frac{16t^8-8t^4+1}{16t^4}\]
Any confusion on how I got that?

Answer 9

nope, I got it!

Answer 10

So I guess that leaves us here,
\[\large \int\limits\limits_1^x \sqrt{1+\frac{16t^8-8t^4+1}{16t^4}}\quad dt\]
Once again, we'll want to get a common denominator between that 1 and the fraction.

Answer 11

\[\large \int\limits\limits\limits_1^x \sqrt{\frac{16t^4}{16t^4}+\frac{16t^8-8t^4+1}{16t^4}}\quad dt\]

Answer 12

\[\large \int\limits\limits\limits\limits_1^x \sqrt{\frac{16t^8+8t^4+1}{16t^4}}\quad dt\]

Answer 13

hmmm okay. Now I just need to find the antidertivative of that?

Answer 14

I have a feeling what we're going to have to do is probably... complete the square in the numerator, and then apply a trig sub.
Have you done work with trigonometric substitutions?

Answer 15

We can take the root of the denominator really nicely. Let's do that, giving us,\[\large \int\limits_1^x \frac{1}{4t^2}\sqrt{16t^8+8t^4+1}\quad dt\]

Answer 16

No I haven't. I think we are skipping that section.

Answer 17

Ah ok, maybe that's not the next step then.
Hmm what to do next. Thinkinggg

Answer 18

factor

Answer 19

under the radical

Answer 20

\[(4t^4+1)^2\]

Answer 21

Oh it's already a perfect square under the radical?
Blah, I dunno why I didn't notice that lol.

Answer 22

\[\large \int\limits\limits_1^x \frac{1}{4t^2}\sqrt{(4t^2+1)^2}\quad dt\]Hmm that works out nicely :)

Answer 23

Do you think you can solve it from here tiffany? :D

Answer 24

yes, thank you both! :)