# Find the exact length of the curve $y = \ln(1-x^2)$

Edit: Update with the full question for context

Find the exact length of the curve $$y = \ln(1-x^2), 0 \leq x \leq \frac{1}{2}$$

The integral below is what I got after finding the derivative $$\frac{-2}{1-x^2}$$ via the chain rule.

Can someone give me a hint on how to evaluate this integral with a range of $$0 \leq x \leq \frac{1}{2}$$? $$I = \int_{0}^{\frac{1}{2}} \sqrt{1+\bigg(\frac{-2x}{1-x^2}\bigg)^2}\mathrm dx$$ More specifically, how to deal with this fraction? That is what I am struggling the most with.

I tried to simplify it into $$\frac{1}{1-2x^2 + x^4} = 1(1-2x^2+x^4)^{-1} = 1 - 2x^{-2} + x^{-4}$$ which gives me

$$I = \int_{0}^{\frac{1}{2}} \sqrt{2 - 2x^{-2} + x^{-4}} \mathrm dx$$ I looked at $$u$$-substitution, where $$u = 2-2x^{-2} + x^{-4}$$, but the $$\mathrm du$$ value didn't work out for me.

I am not entirely sure how trig substitution would work for a fraction//polynomial either.

• I think $$(1-2x^2+x^4)^{-1}\ne1-2x^{-2}+x^{-4}.$$ – awllower Apr 16 at 15:46
• This cannot be integrated in terms of elementary functions. – Peter Foreman Apr 16 at 15:53
• @PeterForeman what does "elementary function" mean? Also dammit I thought power signs had the distributive property – Evan Kim Apr 16 at 15:56
• I mean, you literally will not be able to integrate this without using power series (an approximation to the original function). – Peter Foreman Apr 16 at 15:57
• After your edits the integrand seems to be $\frac{1+x^2}{1-x^2}$ which makes the rest quite straightforward – Golovanov399 Apr 16 at 16:27

\begin{align}I &= \int_{0}^{\frac{1}{2}} \sqrt{1+\bigg(\frac{-2x}{1-x^2}\bigg)^2}\,\mathrm dx\\ &=\int_{0}^{\frac{1}{2}} \frac{\sqrt{(1-x^2)^2+4x^2}}{1-x^2}\mathrm \,dx\\ &=\int_0^\frac12 \frac{\sqrt{1+2x^2+x^4}}{1-x^2}\mathrm\, dx \\ &=\int_0^\frac12 \frac{1+x^2}{1-x^2}\mathrm\, dx \\ &=\int_0^\frac12 \frac{x^2-1+2}{1-x^2}\, dx \\ &= \int_0^\frac12 -1+\frac{2}{1-x^2}\, dx\\ &= \int_0^\frac12 -1 + \frac{1}{1-x} + \frac{1}{1+x} \, dx \\ &= -x - \ln (1-x)+\ln (1+x) |_{0}^{\frac12}\\ &= - \frac12 - \ln \frac12 + \ln \frac32 \\ &= \ln 3 - \frac12 \end{align}