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

The exact length of a curve is a geometrical concept that is addressed by integral calculus. It is a method for calculating the exact lengths of line segments.

## Answer: The exact length of the curve. y = ln (1 − x^{2}), 0 ≤ x ≤ 1/2 is ln (3) - 1/2 units.

Let’s solve it step by step.

**Explanation: **

For the given curve, We will use the formula for the length of the arc(L) of the graph.

Given function ⇒ y = ln (1 – x^{2})

Now, differentiate the given function (y = ln (1 – x^{2})) with respect to “x”

dy/dx = d( ln (1 – x^{2}))/dx

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

For finding arc length (L), we will use the following formula,

L = \(\int_{a}^{b}\) √[1 + (dy/dx)^{2}] dx.

Substituting the value of dy/dx in length of curve formula from (1)

L = \(\int_{a}^{b}\) √[1 + ( - 2x/1 - x^{2})^{2}] dx

= \(\int_{a}^{b} \dfrac{\sqrt{(1-x^2)^2 + 4 x^{2}}}{1-x^2} d x\)

= \(\int_{a}^{b} \dfrac{\sqrt{1+x^4+2 x^{2}}}{1-x^2} d x\)

= \(\int_{a}^{b} \dfrac{1+x^2}{1-x^2} d x\)

Evaluate the indefinite integral

\(\int{ \dfrac{1+x^2}{1-x^2} d x} \)

= \(\int{ \dfrac{x^2 - 1+2 }{1-x^2} d x} \)

= \(\int{ -1 + \dfrac{2 }{1-x^2} d x} \)

Use the property of partial fractions

\(\int{ -1 + \dfrac{1}{1-x}+\dfrac{1}{1+x} d x} \)

= \(-x - \ln|1-x| + \ln|1+x|\)

Now, evaluate the definite integral

\(\int_{0}^{1/2} \dfrac{1+x^2}{1-x^2}dx\)

= \((-x - \ln|1-x| + \ln|1+x|)|_{0} ^{1/2}\)

Calculate the expression

\((-\frac{1}{2}-\ln|1-1/2|+\ln|1+1/2|)-(0-\ln|1-0|+\ln|1+0|)\)

Simplify further

= \(-\frac{1}{2}-\ln(1/2)+\ln(3/2)\)

L ≈ ln (3) - 1/2 units

### Therefore, the length of the curve is ln (3) - 1/2 units.

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