# Need helping solving an integral

• Jul 16th 2009, 04:18 PM
RobLikesBrunch
Need helping solving an integral
Hello, I can't manage to get anything out of this integral:

$\int^4_2 \sqrt{1+(x-\frac{1}{4x})^2}$

I've tried expanding and substituting, as well as factoring out different powers of x's and then substituting, but it never works.

According to my calculus book, it's possible to do it (I'm doing arc length, so you get an abundance of integrals that don't work, but apparently, this one does...somehow)

Ideas?

Note: the original problem gave the equation $y=\frac{x^2}{2}-\frac{ln}{4}, 2\leq x \leq 4$, where it asked me to graph the curve and find its length using the arc length formula.
• Jul 16th 2009, 04:24 PM
NonCommAlg
Quote:

Originally Posted by RobLikesBrunch
Hello, I can't manage to get anything out of this integral:

$\int^4_2 \sqrt{1+(x-\frac{1}{4x})^2}$

I've tried expanding and substituting, as well as factoring out different powers of x's and then substituting, but it never works.

According to my calculus book, it's possible to do it (I'm doing arc length, so you get an abundance of integrals that don't work, but apparently, this one does...somehow)

Ideas?

Note: the original problem gave the equation $y=\frac{x^2}{2}-\frac{ln}{4}, 2\leq x \leq 4$, where it asked me to graph the curve and find its length using the arc length formula.

$1+ \left ( x - \frac{1}{4x} \right)^2 = \left ( x + \frac{1}{4x} \right)^2.$
• Jul 16th 2009, 04:35 PM
RobLikesBrunch
Well that was certainly elegant.

I'm not entirely sure if why that works, though. How come you can simply change the sign by incorporating the 1?

Anyway-- the integral works now. Woopie!
• Jul 16th 2009, 04:41 PM
skeeter
Quote:

Originally Posted by RobLikesBrunch
Well that was certainly elegant.

I'm not entirely sure if why that works, though. How come you can simply change the sign by incorporating the 1?

a bit of algebra ...

$1 + \left(x - \frac{1}{4x}\right)^2$

$1 + \left(x^2 - \frac{1}{2} + \frac{1}{16x^2}\right)$

$x^2 + \frac{1}{2} + \frac{1}{16x^2}$

$\left(x + \frac{1}{4x}\right)^2$