Calculus: Early Transcendentals 8th Edition

The exact length of the curve $$y=\ln(1-x^{2}), \quad\quad 0 \leq x \leq \frac{1}{2}.$$ is equal to $(\ln 3-\frac{1}{2}).$
$$y=\ln(1-x^{2}), \quad\quad 0 \leq x \leq \frac{1}{2}.$$ We have $$y=\ln(1-x^{2}) \Rightarrow d y/ d x=\frac{1}{1-x^{2}}. (-2x)$$ \begin{aligned} 1+\left(\frac{d y}{d x}\right)^{2} &=1+\frac{4 x^{2}}{\left(1-x^{2}\right)^{2}} \\ &=\frac{1-2 x^{2}+x^{4}+4 x^{2}}{\left(1-x^{2}\right)^{2}}\\ &=\frac{1+2 x^{2}+x^{4}}{\left(1-x^{2}\right)^{2}}\\ &=\frac{\left(1+x^{2}\right)^{2}}{\left(1-x^{2}\right)^{2}} \end{aligned} $\Rightarrow$ \begin{aligned} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}& =\sqrt{\left(\frac{1+x^{2}}{1-x^{2}}\right)^{2}} \\ &=\frac{1+x^{2}}{1-x^{2}}\\ & \quad\left[ \text{ by long division } \right]\\ &=-1+\frac{2}{1-x^{2}}\\ & \quad\left[ \text{ its partial fractions are } \right]\\ &=-1+\frac{1}{1+x}+\frac{1}{1-x} \end{aligned} So the arc length is \begin{aligned} L &=\int_{0}^{1 / 2}\left(-1+\frac{1}{1+x}+\frac{1}{1-x}\right) d x \\ &=[-x+\ln |1+x|-\ln |1-x|]_{0}^{1 / 2} \\ &=\left(-\frac{1}{2}+\ln \frac{3}{2}-\ln \frac{1}{2}\right)-0 \\ &=\ln 3-\frac{1}{2} \end{aligned} Hence, the exact length of the curve $$y=\ln(1-x^{2}), \quad\quad 0 \leq x \leq \frac{1}{2}.$$ is equal to $(\ln 3-\frac{1}{2}).$