# Thread: Integration with arc length question

1. ## Integration with arc length question

I have an arc-length problem that has a complicated integral. The problem is: find the length of the curve y=ln(x^2-1) from x=2 to 3.

I found the derivative and squared it, which gave (4x^2)/(x^4-2x^2-1). Then I plugged it into the formula, simplified it until I got the integral of [(x^2+1)^2]/[(x^2-1)^2] from 2 to 3. How do I solve this integral? I know how to set up the integral and simplify it to that point, but I can't get past that point. Any help? Thanks.

2. I derived

$
\int_2^3\,\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,d x=\int_2^3\,\sqrt{1+\left(\frac{2x}{x^2-1}\right)^2}\,dx$

$=\int_2^3\,\sqrt{1+\frac{4x^2}{(x^2-1)^2}}\,dx$
$=\int_2^3\,\sqrt{\frac{(x^2-1)^2}{(x^2-1)^2}+\frac{4x^2}{(x^2-1)^2}}\,dx$
$=\int_2^3\,\sqrt{\frac{x^4-2x^2+1+4x^2}{(x^2-1)^2}}\,dx$
$=\int_2^3\,\sqrt{\frac{x^4+2x^2+1}{(x^2-1)^2}}\,dx$
$=\int_2^3\,\sqrt{\frac{(x^2+1)^2}{(x^2-1)^2}}\,dx$
$=\int_2^3\,\frac{x^2+1}{x^2-1}\,dx.$

The integrand can be decomposed by partial fractions:

$\frac{x^2+1}{x^2-1}=\frac{x^2+1}{(x+1)(x-1)}=\frac{A}{x+1}+\frac{B}{x-1}.$

3. Originally Posted by virtuoso735
I have an arc-length problem that has a complicated integral. The problem is: find the length of the curve y=ln(x^2-1) from x=2 to 3.

I found the derivative and squared it, which gave (4x^2)/(x^4-2x^2-1). Then I plugged it into the formula, simplified it until I got the integral of [(x^2+1)^2]/[(x^2-1)^2] from 2 to 3. How do I solve this integral? I know how to set up the integral and simplify it to that point, but I can't get past that point. Any help? Thanks.
you forgot the square root. what you in fact have is $\int \frac {x^2 + 1}{x^2 - 1}~dx = \int \frac {x^2 - 1 + 2}{x^2 - 1}~dx =$ $\int \left( 1 + \frac 2{x^2 - 1} \right)~dx = \int \left( 1 + \frac 2{(x + 1)(x - 1)} \right)~dx$

now can you finish?

4. Originally Posted by Scott H
The integrand can be decomposed by partial fractions:

$\frac{x^2+1}{x^2-1}=\frac{x^2+1}{(x+1)(x-1)}=\frac{A}{x+1}+\frac{B}{x-1}.$
you can't employ partial fractions decomposition if the degree of the numerator is greater than or equal to the degree of the denominator. you must simplify first, either by long division, or something similar to what i did

5. Thank you, Jhevon. I did not even realize my mistake.