\( \int _{1}^{2}\sqrt {1+(x^{2}-\frac{1}{4x^{2}})^{2}} d x \)

To evaluate the integral \( \int _{1}^{2}\sqrt {1+(x^{2}-\frac{1}{4x^{2}})^{2}} d x \), we can start by simplifying the expression inside the square root. First, let’s simplify \( x^2 - \frac{1}{4x^2} \): \[ x^2 - \frac{1}{4x^2} = \frac{4x^4 - 1}{4x^2} \] Now we square this expression: \[ \left( x^2 - \frac{1}{4x^2} \right)^2 = \left( \frac{4x^4 - 1}{4x^2} \right)^2 = \frac{(4x^4 - 1)^2}{16x^4} \] Next, we place this back into the integral: \[ \sqrt{1 + \left( x^2 - \frac{1}{4x^2} \right)^2} = \sqrt{1 + \frac{(4x^4 - 1)^2}{16x^4}} \] Combining the terms inside the square root: \[ \sqrt{\frac{16x^4 + (4x^4 - 1)^2}{16x^4}} = \frac{\sqrt{16x^4 + (16x^8 - 8x^4 + 1)}}{4x^2} \] This simplifies further to: \[ = \frac{\sqrt{16x^8 + 8x^4 + 1}}{4x^2} \] The integral can be written as: \[ \int_{1}^{2}\frac{\sqrt{(4x^4)^{2} + 2(4x^4)(\frac{1}{2}) + (\frac{1}{2})^{2}}}{4x^2} \, dx \] Next, we find \( 16x^8 + 8x^4 + 1 = (4x^4 + 1)^2 \). Thus, we have: \[ \sqrt{16x^8 + 8x^4 + 1} = 4x^4 + 1 \] We can substitute this back into the integral: \[ \int_{1}^{2} \frac{4x^4 + 1}{4x^2} \, dx = \int_{1}^{2} (x^2 + \frac{1}{4x^2}) \, dx \] Now we can separate this integral: \[ \int_{1}^{2} x^2 \, dx + \int_{1}^{2} \frac{1}{4x^2} \, dx \] Calculating the first integral: \[ \int x^2 \, dx = \frac{x^3}{3} \Big|_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \] Calculating the second integral: \[ \int \frac{1}{4x^2} \, dx = -\frac{1}{4x} \Big|_1^2 = -\left(\frac{1}{8} - \frac{1}{4}\right) = \frac{1}{8} \] Combining these results gives: \[ \frac{7}{3} + \frac{1}{8} = \frac{56 + 3}{24} = \frac{59}{24} \] Thus, the value of the integral is: \[ \int _{1}^{2}\sqrt {1+(x^{2}-\frac{1}{4x^{2}})^{2}} d x = \frac{59}{24} \]