# How do you find the length of the curve for y= ln(1-x²) for (0, 1/2)?

Aug 16, 2017

The formula for the arc length of the curve $y$ on the interval $\left[a , b\right]$ is given by:

$s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Here, where $y = \ln \left(1 - {x}^{2}\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x}{1 - {x}^{2}} = \frac{2 x}{{x}^{2} - 1}$.

Thus, the arc length in question is:

$s = {\int}_{0}^{1 / 2} \sqrt{1 + {\left(\frac{2 x}{{x}^{2} - 1}\right)}^{2}} \mathrm{dx}$

$s = {\int}_{0}^{1 / 2} \sqrt{\frac{{\left({x}^{2} - 1\right)}^{2} + {\left(2 x\right)}^{2}}{{x}^{2} - 1} ^ 2} \mathrm{dx}$

$s = {\int}_{0}^{1 / 2} \frac{\sqrt{{x}^{4} - 2 {x}^{2} + 1 + 4 {x}^{2}}}{{x}^{2} - 1} \mathrm{dx}$

$s = {\int}_{0}^{1 / 2} \frac{\sqrt{{x}^{4} + 2 {x}^{2} + 1}}{{x}^{2} - 1} \mathrm{dx}$

Note this is factorable:

$s = {\int}_{0}^{1 / 2} \frac{\sqrt{{\left({x}^{2} + 1\right)}^{2}}}{{x}^{2} - 1} \mathrm{dx}$

$s = {\int}_{0}^{1 / 2} \frac{{x}^{2} + 1}{{x}^{2} - 1} \mathrm{dx}$

Rewriting:

$s = {\int}_{0}^{1 / 2} \frac{{x}^{2} - 1 + 2}{{x}^{2} - 1} \mathrm{dx}$

$s = {\int}_{0}^{1 / 2} \left(1 + \frac{2}{{x}^{2} - 1}\right) \mathrm{dx}$

Perform partial fractions on the second piece:

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

$2 = A \left(x + 1\right) + B \left(x - 1\right)$

Letting $x = 1$ reveals that

$2 = 2 A \text{ "=>" } A = 1$

And letting $x = - 1$ shows that

$2 = - 2 B \text{ "=>" } B = - 1$

So

$\frac{2}{{x}^{2} - 1} = \frac{1}{x - 1} - \frac{1}{x + 1}$

Then

$s = {\int}_{0}^{1 / 2} \left(1 + \frac{1}{x - 1} - \frac{1}{x + 1}\right) \mathrm{dx}$

Which are all easily integrateable:

$s = {\left[x + \ln \left\mid x - 1 \right\mid - \ln \left\mid x + 1 \right\mid\right]}_{0}^{1 / 2}$

$s = {\left[x + \ln \left\mid \frac{x - 1}{x + 1} \right\mid\right]}_{0}^{1 / 2}$

$s = \left(\frac{1}{2} + \ln \left\mid \frac{\frac{1}{2} - 1}{\frac{1}{2} + 1} \right\mid\right) - \left(0 + \ln \left\mid \frac{0 - 1}{0 + 1} \right\mid\right)$

$s = \frac{1}{2} + \ln \left\mid \frac{- \frac{1}{2}}{\frac{3}{2}} \right\mid + \ln \left\mid - 1 \right\mid$

$s = \frac{1}{2} + \ln \left(\frac{1}{3}\right)$

Or

$s = \frac{1}{2} - \ln 3$