# Math Help - Arc Length Help

1. ## Arc Length Help

Hey guys, I love my Calc. II class and this arc length integral has me kind of stumped. I put the integral in my TI-83 calculator and got an answer but I wanted to see if any of you thought there would be a way to integrate this:

$\int\sqrt{1 + \frac{4}{(2x+1)^2}} dx$ on the interval $[\frac{1}{2},\sqrt{3} - \frac{1}{2}]$

The original problem was calculate the arc length of $y = \ln{2x+1}$ on the interval $[\frac{1}{2},\sqrt{3} - \frac{1}{2}]$

Any ideas? Try to force a perfect square trinomial in there? Thanks for any help!

2. $\left[\frac{d}{dx}(ln(2x)+1)\right]^{2}=\frac{1}{x^{2}}$

Arc length would be $\int_{\frac{1}{2}}^{\sqrt{3}-\frac{1}{2}}\sqrt{1+\frac{1}{x^{2}}}dx$

3. Go to this website Wolfram|Alpha
In the input window, type in this exact expression: integrate sqrt[1 +4/(2x+1)^2]dx from 1/2 to sqrt[3]-(1/2) (you can copy & paste) Click the equals bar at the right-hand end of the input window..

4. Originally Posted by galactus
$\left[\frac{d}{dx}(ln(2x)+1)\right]^{2}=\frac{1}{x^{2}}$

Arc length would be $\int_{\frac{1}{2}}^{\sqrt{3}-\frac{1}{2}}\sqrt{1+\frac{1}{x^{2}}}dx$
Oops.. I put in the code wrong. It's not $y = \ln(2x) + 1$ it's $y = \ln(2x+1)$

Thanks for the Wolfram/Alpha link, I'm going to check it out.

5. Ok so I put the integral into Wolfram Alfa and it gave me the same answer that my calculator gave me when I plugged it into that. Even with the answer that Wolfram Alfa gives me I'm still struggling with how to integrate the problem.

Is there maybe a way I can use trigonometric substitution to get $a\tan\theta$ into the $\frac{4}{(2x+1)^2}$ part of the function?

Thanks for any more help!

6. $\int\sqrt{1+\frac{4}{(2x+1)^{2}}}dx$

If we let $x=\frac{u-1}{2}, \;\ dx=du$

then it whittles down to:

$\int\frac{\sqrt{u^{2}+4}}{u}du$

Now, it is easier to integrate?.

You can try different routes including trig sub.