# What is the arc length of f(x) = x^2-ln(x^2)  on x in [1,3] ?

Jun 28, 2018

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - \ln \left(\frac{4 \sqrt{265} - 55}{45}\right)$ units.

#### Explanation:

$f \left(x\right) = {x}^{2} - \ln \left({x}^{2}\right) = {x}^{2} - 2 \ln x$

$f ' \left(x\right) = 2 x - \frac{2}{x}$

Arc length is given by:

$L = {\int}_{1}^{3} \sqrt{1 + {\left(2 x - \frac{2}{x}\right)}^{2}} \mathrm{dx}$

Expand and simplify:

$L = {\int}_{1}^{3} \frac{\sqrt{4 {x}^{4} - 7 {x}^{2} + 4}}{x} \mathrm{dx}$

For ease of algebra, apply the substitution ${x}^{2} = u$:

$L = \frac{1}{2} {\int}_{1}^{9} \frac{\sqrt{4 {u}^{2} - 7 u + 4}}{u} \mathrm{du}$

Multiply numerator and denominator by $\sqrt{4 {u}^{2} - 7 u + 4}$

$L = \frac{1}{2} {\int}_{1}^{9} \frac{4 {u}^{2} - 7 u + 4}{u \sqrt{4 {u}^{2} - 7 u + 4}} \mathrm{du}$

Rearrange so that the numerator contains the derivative of the denominator:

$L = \frac{1}{4} {\int}_{1}^{9} \frac{\left(8 u - 7\right) - \left(7 - \frac{8}{u}\right)}{\sqrt{4 {u}^{2} - 7 u + 4}} \mathrm{du}$

Integration is distributive:

$L = \frac{1}{4} {\int}_{1}^{9} \frac{8 u - 7}{\sqrt{4 {u}^{2} - 7 u + 4}} \mathrm{du} - \frac{1}{4} {\int}_{1}^{9} \frac{7 - \frac{8}{u}}{\sqrt{4 {u}^{2} - 7 u + 4}} \mathrm{du}$

Complete the square in the square root:

$L = \frac{1}{2} {\left[\sqrt{4 {u}^{2} - 7 u + 4}\right]}_{1}^{9} - {\int}_{1}^{9} \frac{7 - \frac{8}{u}}{\sqrt{{\left(8 u - 7\right)}^{2} + 15}} \mathrm{du}$

Apply the substitution $8 u - 7 = \sqrt{15} \tan \theta$:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \int \frac{\left(7 - \frac{64}{\sqrt{15} \tan \theta + 7}\right)}{\sqrt{15} \sec \theta} \left(\frac{\sqrt{15}}{8} {\sec}^{2} \theta d \theta\right)$

Simplify and distribute:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \int \sec \theta d \theta + 8 \int \sec \frac{\theta}{\sqrt{15} \tan \theta + 7} d \theta$

Apply the appropriate double-angle Trigonometric identities:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \left[\ln | \sqrt{15} \sec \theta + \sqrt{15} \tan \theta |\right] + 8 \int {\sec}^{2} \frac{\frac{\theta}{2}}{2 \sqrt{15} \tan \left(\frac{\theta}{2}\right) + 7 \left(1 - {\tan}^{2} \left(\frac{\theta}{2}\right)\right)} d \theta$

Rearrange:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} {\left[\ln | \sqrt{{\left(8 u - 7\right)}^{2} + 15} + 8 u - 7 |\right]}_{1}^{9} - 8 \int {\sec}^{2} \frac{\frac{\theta}{2}}{7 {\tan}^{2} \left(\frac{\theta}{2}\right) - 2 \sqrt{15} \tan \left(\frac{\theta}{2}\right) - 7} d \theta$

Complete the square in the denominator:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - 56 \int {\sec}^{2} \frac{\frac{\theta}{2}}{{\left(7 \tan \left(\frac{\theta}{2}\right) - \sqrt{15}\right)}^{2} - 64} d \theta$

Apply the difference of squares:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - 56 \int {\sec}^{2} \frac{\frac{\theta}{2}}{\left(7 \tan \left(\frac{\theta}{2}\right) - \sqrt{15} - 8\right) \left(7 \tan \left(\frac{\theta}{2}\right) - \sqrt{15} + 8\right)} d \theta$

Apply partial fraction decomposition:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - \frac{7}{2} \int \left(\frac{1}{7 \tan \left(\frac{\theta}{2}\right) - \sqrt{15} - 8} - \frac{1}{7 \tan \left(\frac{\theta}{2}\right) - \sqrt{15} + 8}\right) {\sec}^{2} \left(\frac{\theta}{2}\right) d \theta$

Integrate term by term:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - \left[\ln | 7 \tan \left(\frac{\theta}{2}\right) - \sqrt{15} - 8 | - \ln | 7 \tan \left(\frac{\theta}{2}\right) - \sqrt{15} + 8 |\right]$

Apply the appropriate half-angle Trigonometric identity:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - \left[\ln | \frac{7 \sqrt{15} \tan \theta - \left(\sqrt{15} + 8\right) \left(\sqrt{15} + \sqrt{15} \sec \theta\right)}{7 \sqrt{15} \tan \theta - \left(\sqrt{15} - 8\right) \left(\sqrt{15} + \sqrt{15} \sec \theta\right)} |\right]$

Reverse the last substitution:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - {\left[\ln | \frac{7 \left(8 u - 7\right) - \left(\sqrt{15} + 8\right) \left(\sqrt{15} + \sqrt{{\left(8 u - 7\right)}^{2} + 15}\right)}{7 \left(8 u - 7\right) - \left(\sqrt{15} - 8\right) \left(\sqrt{15} + \sqrt{{\left(8 u - 7\right)}^{2} + 15}\right)} |\right]}_{1}^{9}$

Insert the limits of integration:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - \ln | \frac{455 - \left(\sqrt{15} + 8\right) \left(\sqrt{15} + 4 \sqrt{265}\right)}{455 - \left(\sqrt{15} - 8\right) \left(\sqrt{15} + 4 \sqrt{265}\right)} \cdot \frac{7 - \left(\sqrt{15} - 8\right) \left(\sqrt{15} + 4\right)}{7 - \left(\sqrt{15} + 8\right) \left(\sqrt{15} + 4\right)} |$

Simplify:

$L = \frac{1}{2} \left(\sqrt{265} - 1\right) - \frac{7}{8} \ln \left(\frac{4 \sqrt{265} + 65}{5}\right) - \ln \left(\frac{4 \sqrt{265} - 55}{45}\right)$