# Math Help - [SOLVED] Arc length integral

1. ## [SOLVED] Arc length integral

Hey, I derived an integral for a student, but can't find a way to integrate it.

$\int \sqrt{1 + \left ( \frac{3x^2}{4} - \frac{1}{3x^2} \right )^2}~dx$

My calculator gives
$\frac{3x^4 - 4}{12x} + C$

Any ideas how to do it?

Thanks.
-Dan

2. $\int {\sqrt {1 + \frac{{9x^4 }}
{{16}} - \frac{1}
{2} + \frac{1}
{{9x^4 }}} \,dx} = \int {\sqrt {\frac{{9x^4 }}
{{16}} + \frac{1}
{2} + \frac{1}
{{9x^4 }}} }\,dx = \int {\left( {\frac{{3x^2 }}
{4} + \frac{1}
{{3x^2 }}} \right)\,dx} .$

And we're done.

3. Originally Posted by Krizalid
$\int {\sqrt {1 + \frac{{9x^4 }}
{{16}} - \frac{1}
{2} + \frac{1}
{{9x^4 }}} \,dx} = \int {\sqrt {\frac{{9x^4 }}
{{16}} + \frac{1}
{2} + \frac{1}
{{9x^4 }}} }\,dx = \int {\left( {\frac{{3x^2 }}
{4} + \frac{1}
{{3x^2 }}} \right)\,dx} .$

And we're done.
Oh dear God.

It was right in front of me the whole time.

Thanks!

-Dan