Question

find the exact length of the curve

y = ln(1-x^2), 0 <= x <= (1/2)

So by doing the work I eventually get down to

integral from 0 to 1/2 (1 + x^2)/(1-x^2) dx

but I keep getting

-1 + 1/(x-1) - 1/(x+1) after partial fractions which gives the answer

ln(1/3) - 1/2 when I evaluate the integral

the answer is ln(3) - 1/2 and it seems I am screwing up during the partial fraction. What exactly is wrong here?

Answer 1

the length of curve
= integral [ √(1 + (dy/dx)^2) from 0 to 1/2

dy/dx = -2x/(1-x^2) = 2x/(x^2-1)
which I separated into
1/(x+1) + 1/(x-1) using partial fractions.

we now have to square this, add it to one and take the square root.
Perhaps you tried to split it into partial fractions after you squared the derivative, I did it before.

integral [ √( 1 + ( 1/(x+1) + 1/(x-1) )^2 ) ]

does that help so far?

Wolfram says this:
http://integrals.wolfram.com/index.jsp?expr=√%281+%2B+%282x%2F%28x%5E2-1%29%29%5E2%29&random=false