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# Find the exact length of the curve. y = ln(1 − x^{2} ), 0 ≤ x ≤ 1/9.

**Solution:**

Given, y = ln(1-x^{2} ), 0 ≤ x ≤ 1/9

Length of the curve, y = f(x) from x =a tox = b is given by:

\( \int_{a}^{b}\sqrt{1+f'(x)^{2}} .dx \)

y = ln(1-x^{2})

Differentiating w.r.t. x,

dy/dx = 1/(1 - x^{2}) × (-2x) = -2x/(1 - x^{2})

Length of the curve,

= \( \int_{0}^{1/9}\sqrt{1+\left [\frac{-2x}{(1-x^{2})} \right ]^{2}} .dx \)

= \( \int_{0}^{1/9}\sqrt{\left [ \frac{1 +x^{4}-2x^{2}+4x^{2}}{(1-x^{2})^{2}} \right ]} .dx \)

= \( \int_{0}^{1/9}\sqrt{\frac{(1+x^{2})^{2}}{(1-x^{2})^{2}}} . dx \)

= \( \int_{0}^{1/9}\frac{1+x^{2}}{1-x^{2}} . dx \)

{By dividing numerator by denominator and expressing using

Dividend/Divisor = Quotient + (Remainder/Divisor)]}

= \( \int_{0}^{1/9}-1 + \frac{2}{1-x^{2}}.dx \)

= \( -x\Biggr|_{0}^{1/9} + 2 \frac{1}{2} log\left | \frac{1 + x}{1-x} \right |\Biggr|_{0}^{1/9} \)

= \( -\left ( \frac{1}{9}-0 \right ) + log \left | \frac{1+ \frac{1}{9}}{1-\frac{1}{9}} \right |-log\left | \frac{1+0}{1-0} \right | \)

= \( -\frac{1}{9} + log\left | \frac{10}{8} \right |- log (1) \)

= \( -\frac{1}{9} + log\left | \frac{5}{4} \right |- 0 \)

= log(5/4) - 1/9

## Find the exact length of the curve. y = ln(1 − x^{2} ), 0 ≤ x ≤ 1/9.

**Summary:**

The exact length of the curve, y = ln(1-x^{2}), 0 ≤ x ≤ 1/9 is log(5/4) - 1/9.

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