Question

How do I integrate this function
(1+((x/4)-(1/x))^2)^(1/2)

Answer 1

\[\int\limits_{?}^{?} \sqrt{1+(\frac{ x }{ 4 }-\frac{ 1 }{ x }})^2\]

Answer 2

\[\int \left[\left(1+ \left(\frac{x}{4}-\frac{1}{x}\right)\right)^2\right]^{\frac{1}{2}}dx\]

Answer 3

Mmm this one is kinda fun.
Bunch of messy algebra to get through

Answer 4

or yeah.... use the square root. Hah.

Answer 5

It's an arc length problem, if that matters at all.

The original equation is y=(1/8)x^2 -ln(x) and we're supposed to find arc length of this.

Answer 6

ohhh

Omg I think I just figured it out

Answer 7

\[\sqrt{1+ (\frac{ x }{ 4 }-\frac{ 1 }{ x })^{2}} = \sqrt{1+ (\frac{ x^{2}-4 }{ 4x })^{2}} = \sqrt{1 + \frac{ x^{4} - 8x^{2} + 16 }{ 16x^{2} }} \]
\[= \sqrt{\frac{ 16x^{2} + x^{4} - 8x^{2} + 16 }{ 16x^{2} }} = \frac{ \sqrt{x^{4} + 8x^{2} + 16} }{ \sqrt{16x^{2}} } = \frac{ \sqrt{(x^{2} + 4)^{2}} }{ 4x } = \frac{ x^{2} + 4 }{ 4x }\]

Answer 8

Arc length formula as defined: \[ L = \int_{a}^{b} \sqrt{1+(h'(y))^2}dy = \int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2}dy\] Just for reference.

Answer 9

Can actually look at the integral and tell it's an arc length problem without you even mentioning it :) A lot of the time, the simplication of the square root you get is the EXACT same as the derivative of your function but with a change of sign. And as you notice, that's what we have. If you simplify the (x^2+4)/4x I have, it turns into your derivative, but plus sign instead of a minus. This happens A LOT.

So if you're doing these arc length problems and you're unsure how to simplify it, check and see if the simplification of the square root is simply your derivative but with a change of sign.

Answer 10

What about this?

Foil out the parenthesis and you get

\[\sqrt{\frac{ 1 }{ 2 }+\frac{ 1 }{ 16 }x^2+\frac{ 1 }{ x^2 }}\]

Then factor this into a perfect square

\[\sqrt{(\frac{ 1 }{ 4x }+\frac{ 1 }{ x })^2}\]

That seems to work

Answer 11

The X part is supposed to be in the numerator for the first fraction

Answer 12

Yeah, that works out in this case. I'm just used to needing to do more fancy stuff with it to get it to work. Either you end up with the original derivative with a sign change or you end up where you have to factor something out of the square root in order to force a u-substitution. But hey, ya got it :) Now just integrate that with whatever your limits are and you're good.

Answer 13

Cool thank you. I've been struggling with integrals in this structure over and over again, glad I finally figured it out.

Answer 14

No problem :)