<s>[INST] Calculating the length of a curve for $ y=\ln (1-x^2) $ In calculating the  length   of $ y=\ln (1-x^2) $ on the interval $ [0, {3\over4}]$, I found that for whatever reason, I end up with invalid domains in the indefinite integral using the given limits of integration.
Using the formula $ \int^b_a \sqrt{1+f'(x)^2} $, I was able to derive
$$ {dy\over dx}=f'(x)=-{2x\over1-x^2}  $$
$$ \int^{3\over4}_0 \sqrt{1+(-{2x\over1-x^2})^2}\ dx = \int^{3\over4}_0 \sqrt{1+{4x^2\over1-2x^2+x^4}}\ dx = \int^{3\over4}_0 \sqrt{{x^4+2x^2+1}\over{x^4-2x^2+1}}\ dx $$
$$ = \int^{3\over4}_0 {{x^2+1}\over{x^2-1}}\ dx $$
as the integrand in question. Then by the rule of the sums of integrals:
$$ \int^{3\over4}_0 {{x^2}\ dx\over{x^2-1}} + \int^{3\over4}_0 {{dx}\over{x^2-1}}$$
The clearest line of attack was through the following trigonometric substitution:
$$ x=\sec \theta $$
$$ dx=\sec \theta \tan \theta\ d\theta$$
$$ \int {{\sec^3 \theta \tan \theta\ d\theta}\over{\tan^2 \theta}} + \int {{\sec \theta \tan \theta}\over{\tan^2 \theta}} = \int \sec^2 \theta \csc \theta\ d\theta + \int \csc \theta\ d\theta$$
The former I integrated by parts:
**I.**$$ u=\csc \theta\ \quad dv=\sec^2 \theta\ d\theta$$ 
$$ du=-\csc \theta \cot \theta\ d\theta\ \quad v=\tan \theta$$
$$ \int \sec^2 \theta \csc \theta\ d\theta = \boxed{\csc \theta \tan \theta + \int \csc \theta\ d\theta}$$
And the latter:
**II.**$${\int \csc \theta\ d\theta= \boxed{-\ln \lvert{\csc \theta + \cot \theta}\rvert}}$$
Hence the complete indefinite integral:
**I.+II.**$$ \int \sec^2 \theta \csc \theta\ d\theta + \int \csc \theta\ d\theta = \boxed{\csc \theta \tan \theta -2 \ln \lvert{\csc \theta + \cot \theta}\rvert}$$
Transforming back to the original variable:
$$ x=\sec \theta \quad\quad \sqrt{x^2-1} = \tan \theta$$
$$ {1\over\sqrt{x^2-1}} = \cot \theta \quad\quad {x\over\sqrt{x^2-1}} = \csc \theta$$
$$ {\csc \theta \tan \theta -2 \ln \lvert{\csc \theta + \cot \theta}\rvert} = \boxed{x - 2\ln \lvert{{x+1}\over{\sqrt{x^2-1}}}\rvert} $$
As one can easily see, for whatever reason the limits produce a negative term under the radical in the natural log. Any insight as to why this is the case would be much appreciated.
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The hitch is here :

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